University  of  California  •  Berkeley 

The  Theodore  P.  Hill  Collection 

of 
Early  American  Mathematics  Books 


GRADATIONS 


IN 


ALGEBRA, 


IN  WHICH 


THE   FIRST   PRINCIPLES  OF   ANALYSIS 


INDUCTIVELY  EXPLAINED. 


ILLUSTRATED  BY 


COPIOUS  EXERCISES, 


AND  MADE  SUITABLE  FOR  PRIMARY  SCHOOLS. 


BY 

RICHARD  W.  GREEN,  A.  M., 

AUTHOR  OF  ARITHMETICAL  GUIDE,  LITTLE  RECKONER,  ETC. 


PHILADELPHIA : 

PRINTED  BY  I.  A8HMEAD  AND  CO. 
1839. 


Entered  according  to  the  act  of  Congress,  in  the  year  1839,  by 

RICHARD  W.  GREEN, 

in  the  Clerk's  Office  of  the  District  Court  of  the  Eastern  District 
of  Pennsylvania. 


CONTENTS 


Preface, 


NUMERAL  ALGEBRA. 

Preliminary  Remarks,          -                -                -  -           13 

Addition  and  Subtraction  of  Simple  Quantities,     .    -  19^ 

General  Rule  for  uniting  Terms,        -                -  -          21 

Multiplication  and  Division  of  Simple  Quantities,     -  23 

Simple  Equations,               -                -                -  -          24 

I.  Equations  Solved  by  uniting  Terms,  -  -  25 
Addition  of  Compound  Quantities,.  -  -  -  30 
Transposition,               ....  32 

II.  Equations  Solved  by  Transposition,  -  -  32 
Transposition  by  Addition,          ...  35 

III.  Equations,     -                 -                 -                 -  -           36 

Transposition  of  the  Unknown  Quantity,  -                -  38 

IV.  Equations,     -                -                -                -  -39 

Multiplication  of  Compound  Quantities  by  Simple  Quantities,    43 

V.  Equations,      -                 -                 -                 -  -           44 

Fractions,     -                -                -                -                -  49 

Equivalent  Fractions,              -            -                -  -           52 

VI.  Equations,  ....  54 
Division  of  Compound  Quantities  by  Simple  Quantities,-  59 
VIL  Equations,            ....  60 


IV  CONTEXTS. 

Division  of  Fractions  and  Fractions  of  Fractions,  -          63 

VIII.  Equations,           -                -                -                -  66 

Subtraction  of  Compound  Quantities,                -  -          68 

IX.  Equations,              -                 -                 .                 -  70 

Uniting  Fractions  of  different  denominators,  -          t2 

Ratio  and  Proportion,                   ...  76 

X.  Equations,      -                -                -                -  -          78 

Xr.  Equations,             ....  80 

Equations  with  two  Unknown  Quantities,  first  method,  83 

First  method  of  Extermination,          -                -  -          84 

XII.  Equations,            -                -                .                -  88 
Second  method  of  Extermination,      -                -  -          93 

XIII.  Equations,           ...                -  94 

Third  method  of  Extermination,        -                -  -          97 

XIV.  Equations,           ....  98 

Equations  with  several  Unknown  Quantities,  -        102 


LITERAL  ALGEBRA. 


General  Principles,  -  -  - 

Addition  and  Subtraction  of  Algebraic  Quantities, 
Multiplication  of  Algebraic  Quantities, 
General  Properties  of  Numbers, 
Division  of  Algebraic  Quantities, 
Reduction  of  Fractions  to  lower  Tei-ms,     - 
Multiplication  where  one  factor  is  a  fraction. 
Division  of  Fractions,  -  -  - 

Fractions  of  Fractions,         ... 
Uniting  Fractions  of  different  denominators, 
Division  by  Fractions,          ... 
Reducing  Complex  Fi-actipns  tq  Simple  ones, 


104 
110 
113 
120 
123 
131 
133 
135 
139 
141 
143 
146 


PREFACE. 

Thb  object  of  the  author,  in  composing  this  treatise,  was 
to  form  an  easy  introduction  to  the  first  principles  of  alge- 
braical reasoning ;  and  also  to  embrace,  in  the  same  course, 
a  popular  exposition  of  the  most  important  elements  of 
arithmetic.  And  he  believes  that  he  has  been  enabled  to 
combine  the  rudiments  of  both,  in  such  a  manner  as  to  make 
the  operations  of  one  illustrate  the  principles  of  the  other. 

In  order  that  this  method  of  treating  the  subject  might  pre- 
serve its  chief  advantage,  especially  in  the  initiatory  course 
of  the  study  ;  the  work  has  been  divided  into  two  parts — 
Numeral  Algebra  and  Literal  Algebra. 

In  Numeral  Algebra  I  have  treated  of  the  several  arithme- 
tical operations;  first  making  them  intelligible  to  very  young 
pupils,  and  then  exhibiting  them  under  the  algebraical  nota- 
tion. By  this  means,  as  every  lesson  in  algebra  is  immediate- 
ly preceded  by  corresponding  numerical  exercises,  the  tran- 
sition from  one  to  the  other  has  been  made  so  trifling,  that 
the  pupil  will  feel  at  each  step  that  he  has  met  with  nothing 
more  than  what  he  has  already  made  himself  familiar  with, 
in  a  diflierent  dress.  Besides,  as  algebraical  operations  re- 
quire the  exercise  of  abstraction  in  a  greater  degree  than  the 
pupil  is  supposed  to  be  accustomed  to,  I  have  taken  care 
that  the  exercise  on  each  of  the  fundamental  rules,  shall  be 
followed  by  a  selection  of  problems  to  be  solved  by  equa- 
tions. 

As  mathematical  questions  of  this  kind  are  always  pleas- 
ing to  young  pupils,  this  arrangement  will  serve  to  impart 
an  interest  to  the  study  at  the  commencement,  and  also  to 
preserve  a  taste  for  it  through  the  whole  work. 

Under  the  head  of  Literal  Algebra,  I  have  repeated,  in 


TI  PKEFACTET. 

a  more  strictly  algebraical  form,  the  principles  which  have 
been  explained  in  the  preceding  part  of  the  work;  and  have 
shown  some  of  their  uses  by  applying  them  in  the  deduc- 
tion and  demonstration  of  several  abstruse  operations  on 
numbers* 

But  the  great  peculiarity  of  the  book  is,  that  it  habituates 
the  speech  and  the  ear  to  mathematical  language.  In  any 
study,  it  is  necessary  for  beginners  to  receive  such  a  course 
of  training  as  will  imprint  upon  their  minds  each  new  idea, 
as  soon  as  it  is  apprehended.  Learners  in  the  mathematics 
especially,  are  accustomed  to  forget  soon,  both  the  names 
and  the  use  of  the  signs ;  and  also  the  arrangement  of  the 
several  steps  in  the  solution  of  their  problems.  On  this 
account  I  have  required  the  pupil  always  ta  repeat  verbally 
the  operation  that  he  has  performed  ;  taking  care  to  omit  no 
part  of  the  work  that  would  hinder  an  auditor  from  under- 
standing the  reason  for  the  several  steps,  and  consenting  to 
the  just  conclusions  of  the  answer  which  has  been  obtained. 
It  has  been  found  by  experience  that  this  simple  device  ena- 
bles the  young  pupil  to  acquire  the  science  very  easily;  and 
while  it  impresses  his  lessons  indelibly  upon  his  memory,, 
it  also  developes  his  genius,  rectifies  his  inventive  faculties, 
and  imparts,  as  it  were,  a  mathematical  form  to  his  mind ; 
so  much  so,  that  he  is  generally  capable  of  pursuing  the  sub- 
ject afterwards  by  himself. 

In  order  to  accomplish  this  end  more  perfectly,  I  have 
swelled  the  number  of  examples  beyond  the  ordinary  limits. 
These  should  be  thoroughly  mastered  as  the  pupil  proceeds. 
There  must  be  no  smattering  in  the  beginning  of  a  science 
if  the  learner  is  to  continue  the  study.  The  author  has 
found  by  long  experience,  that  a  book  is  sooner  finished 
when  each  part  has  been  made  familiar  to  the  mind,  than 
when  it  has  been  superficially  attended  to. 


PREFACE.  Vll 

With  regard  to  the  arrangement  of  the  several  divisions, 
I  have  been  careful  to  introduce  first  those  prindples  that 
will  be  the  most  easily  apprehended ;  and  afterwards  such 
others  as  would  most  naturally  arise  from  the  former  if  the 
study  vvei-e  entirely  new.  This  method  appears  to  be  tlie 
best  adapted  for  teaching  the  rudiments  of  a  science ; 
although  in  a  succeeding  text  book,  it  Is  necessary  that  the 
arrangement  of  the  several  parts  should  be  more  systematic. 
On  this  account  the  advanced  scholar  must  not  be  surprised 
to  find  in  the  middle  of  the  book,  what  he  has  been  accus- 
tomed to  see  near  the  beginning  of  other  treatises.  How- 
ever, so  much  regard  to  a  regularity  of  arrangement  has 
been  attended  to,  that  the  pupil  will  be  assisted  by  the  asso- 
ciations of  method,  both  to  understand  and  to  remember. 

As  the  author  wishes  to  bring  the  study  of  Algebra  within 
the  reach  of  common  schools,  he  has  endeavored  to  pre- 
pare this  work,  so  that  it  may  be  studied  by  pupils  who  are 
not  already  adepts  in  arithmetic.  And  it  is  believed  that 
such  learners  will  not  fail  of  obtaining,  by  a  perusal  of  it, 
a  full  understanding  of  vulgar  fractions,  roots  and  powers, 
proportion,  progression,  and  other  numerical  operations  that 
are  generally  embraced  in  arithmetical  treatises. 


ADVERTISEMENT. 

The  foregoing  is  tlie  Preface  of  the  author's  "  Inductive 
Treatise  on  the  Elementary  Principles  of  Algebra."  The 
first  146  pages  of  that  book  have  been  published  in  this  form, 
in  order  to  afford  a  cheap  manual  for  those  classes  that  do 
not  wish  to  study  beyond  Simple  Equations.  In  the  pre- 
sent state  of  education,  so  much  of  Algebra  should  be  studied 
by  every  pupil  in  our  common  schools. 

R.  W.  G. 


« 

31 

« 

u 

35 

u 

(( 

38 

(( 

(( 

41 

(( 

ERRATA. 

Before  the  pupil  begins  to  perform  the  sums  in  this  work, 
he  is  requested  to  make  the  following  corrections : 

Page    17  line  15  erase  comma  in  130. 

"      43    "    11  instead  o^  itself  j  read  the  multiplier. 
«      26    "      2  for  36x,  read  35a-. 

6  for  173,  read  17z. 

6  for  76,  read  66. 

in  §54,  for  32,  read  37. 

9,  for  24  read  20. 
"      45  example  5,  for  181,  read  367. 
"       "         **       9,  for  -jV,  read  1|. 
"      48  line    1 ,  for  30  read  80. 
"      55    "    10  insert  f. 

"      65  example  13,  for  14a;— 63,  read  14a;+63. 
«      66         "         5,  for  273,  read  213. 
«'      67         "         5,  for  1000,  read  2280. 

"      70         "       14,  for  7  —  ^,  read  1=?^. 

o  3 

23?  23? 

«      74        «         4,  for  —  —,  read  +-^, 

"      75        «         8,  for  3+?^,  read  ^±^. 

9  9 

"       «  «       10,  for  6  +  2y,  read  3  +  2i/. 

"      77  line  24,  for  23,  read  33. 

"      86  example   6,   for   — 3y   and    — 6y,   read    -{-Sy 

and  -\-6y. 

"      90  example    8,  in  ans.  for  4«,  read  6«. 

*'      91       "        12,  for  35,  read  31. 

"      93  line  4,  for  ^,  read  ^. 

"      97    "    7,  for  10,  read  16. 

*'  112  example  1,  for  4a:,  read  Aax. 


PART   L 


NUMERAL  ALGEBRA. 

PRELIMINARY    REMARKS. 

§1.  Algebra  is  a  method  of  computation,  in  which  we 
find  the  value  of  an  unknown  quantity,  by  using,  in  our 
calculations,  that  quantity  itself* 

Example  1.  A  friend  has  told  me,  that  in  both  of  his 
vest  pockets  he  has  28  cents ;  but  in  the  right  pocket  there 
are  8  cents  more  than  in  the  left. — I  will  use  the  following 
method  of  finding  how  many  cents  there  are  in  each  pocket. 

By  the  statement,  the  right  pocket  contains  as  much  as 
the  left  pocket  and  8  cents  besides. 

Therefore,  instead  of  saying,  28  cents  is  as  much  as  in 
the  left  pocket  and  also  in  the  right  pocket;  we  may  say  28 
cents  is  as  much  as  in  the  left  pocket,  and  again  in  the  left 
pocket,  and  8  cents  besides. 

Hence,  28  cents  is  as  much  as  there  is  in  the  left  pockety 
tvnce,  and  8  cents  besides.     And,  as  this  is  the  case, 

27  cents  is  as  much  as  twice  the  left  pocket  and  7  cents. 

26  cents  is  as  much  as  twice  the  left  pocket  and  6  cents. 
And  so  on,  making  the  cents  fewer  and  fewer;  at  last, 
20  cents  is  as  much  as  twice  the  left  pocket. 

Therefore  10  cents  is  as  much  as  once  the  left  pocket; 
that  is,  there  is  10  cents  in  the  left  pocket.     Of  course  there 
is  18  cents  in  the  right  pocket. 
2 


14  ALGEBRA. 

Example  2.  A  brother  told  his  sister,  that  he  was  4  years 
older  than  she ;  and  that  his  age  and  her  age  put  together, 
made  18  years.     The  age  of  each  may  be  found  as  follows. 

By  the  statement,  his  age  is  the  same  as  her  age  and  4 
years  besides.  Therefore,  instead  of  saying  18  years  is  as 
much  as  her  age  and  his  age  put  together ;  we  can  say 

18  years  is  as  much  as  her  age,  and  also  her  age  and  4 
years  more. 

But  her  age,  and  her  age,  is  the  same  as  twice  her  age. 

Therefore,  18  years  is  as  much  as  twice  her  age  and  4 
years  more.     And  by  taking  away  the  4  years, 
14  years  is  as  much  as  twice  her  age. 
Therefore  7  years  is  once  her  age;  and  as  he  is  4  years 
older,  his  age  is  11  years. 

§2.  The  method  in  which  the  two  preceding  sums  were 
performed,  was  in  use  for  many  years.  But  in  course  of 
time,  mathematicians  began  to  abbreviate  the  words  that 
were  made  use  of  in  their  operations.  We  will  proceed  to 
show  a  few  of  these  abbreviations. 

§J^.  About  the  year  1554,  Stifelius,  a  German,  introduced 
the  sign  +  for  the  words  added  to;  and  called  it  plus. 
Ever  since  that  time,  -f  has  been  used  to  signify  that  the 
«  quantity  after  it,  has  been  added  to  the  quantity  before  it. 
Thus,  24-.6,  is  read  2  plus  6';  and  is  the  same  as,  2  with 
6  added  to  it. 

Example  3.  Divide  $1000  among  A,  B,  and  C ;  so  that 
B  shall  have  $72  more  than  A,  and  C  shall  have  f  172 
•        more  than  A. 

J  By  the  statement,  B's  share  is  equal  to  A's  and  $72  be- 

sides.    We  may  write  it  thus,  ./2's+$72. 

C's  share  is  equal  to  A's  and  $172  besides.  We  may 
write  his  share  thus,  A's+$172. 

Then  all  three  shares  put  together,  will  be  written,  A^s, 


PRELIMINARY  REMARKS.  15 

and  il's+872,  and  A's-\-$l72;  ov,  A's-\-A's  +  $72-i-A's 
+8172. 
Therefore,    A's+A's+$72+A's+8l72,    is    equal    to 

$1000. 
Putting  together  the  $72  and  $172, 

A's+A'5+A's+8244,  is  equal  to  $1000. 
Then,  3  times  A'5+S244,  is  equal  to  $1000. 

Taking  away  $244,     3  times  A'5,  is  equal  to  $756. 
Of  course,  once  A^s  share  is  equal  to  -i  of  $756  ; 

which  is  $252. 

B's  share  is  $252 +$72  ;  which  is  $324. 

C's  share  is  $252 +$172  ;  which  is  $424. 


Proof,  $1000. 

§4.  The  student  will  see  that  in  the  operation  of  the  last 
sum,  instead  of  writing  A^s  share^  we  have  written  simply 
A^s.  This  method  of  abbreviation  was  used  for  for  a  long 
time  before  Stifelius  wrote.  But  he  simplified  the  operation, 
by  employing  only  some  capital  letter  for  every  unknown 
number. 

Example  4.  My  purse  and  money  together,  are  worth  40 
shillings ;  and  the  money  is  7  times  the  worth  of  the  purse. 
How  much  money  is  in  the  purse? 

Now  instead  of  writing  the  worth  of  the  purse,  we  will 
represent  it  by  the  capital  P;  which  may  well  stand  for 
Purse.     Then  the  money  will  be  as  much  as  7  tunes  P. 

Therefore,  P+7  times  P,  is  as  much  as  40s.     Putting 
the  P^s  together,  8  times  P  is  equal  to  40s.   And  of  course. 
Once  P  15  |-  of  40s ;  which  is  5s. 
The  money  is  7  times  5s  ;  which  is  35s. 


Proof,  40s. 
§5.  About  the  year  1550,  John  Scheubelius,  a  Grerman, 


16  ALGEBRA. 

introduced  the  following  practice.  Instead  of  writing  2 
times  A,  or  3  times  A,  or  4  times  B,  &c.;  he  wrote  2  A,  3 
A,  4  B,  &c. 

Example  5.  Two  men,  A  and  B,  owe  me  $270;  and  B's 
debt  is  twice  as  much  as  A's.  How  much  does  each  of 
them  owe  me? 

We  will  represent  A's  debt  by  A ;  and  then  B*s  debt  will 
be  twice  as  much,  or  2  A. 

But  both,  when  put  together,  are  equal  to  8270. 
Therefore,  A+2A,  is  equal  to  $270. 

Putting  the  A's  together,         3A,  is  equal  to  $270. 

Then  A's  debt  is  ^  of  $270  ;  which  .s  $90. 

B's  debt  is  twice  as  much;  which  is  $180. 


Proof,  $270. 

§6.  In  1557,  Dr.  Recorde,  an  Englishman,  introduced 

the  sign  =,  which  we  call  equals.     It  is  used  instead  of 

the  expression,  is  equal  to,  or,  is  as  much  as.     Thus,  2-f6 

=8,  is  read,  2  plus  6,  equals  8. 

Example  6.  A's  age  is  double  of  B's;  and  B's  age  is 
three  times  C's ;  and  the  sum  of  all  their  ages  is  70  years. 
What  is  the  age  of  each? 

Let  us  represent  C's  age  by  the  capital  C.  Then  B's  age 
will  be  three  times  as  much ;  that  is  3C.  And  A's  age  will 
be  twice  that,  which  is  6C.  Then  all  of  them  will  be, 
C-f3C  +  6C.     Therefore 

04-30+60=70. 
Putting  the  O's  together,  100=70. 

Dividing  by  10,  0=7. 

C's  age=7  years. 
i?'s  is  three  times  as  much,  or  21  years. 
ji''s  is  twice  B^s;  or  42  years. 

^t^^f  Tf?*^  Proof,  70 


PRELIMINARY  REMARKS.  17 

§7.  We  very  often  wish  to  state  that  we  have  made  a 
quantity  to  be  less.  This  was  formerly  done  by  using  the 
word  MINUS.  Thus,  if  I  wished  to  say  that  your  age  is  6 
years  less  than  mine ;  I  may  say  your  age  is  equal  to  mine 
when  6  years  are  subtracted  from  mine ;  or,  in  fewer  words, 
your  age  is  equal  to  my  age  minus  6  years. 

§8  Stifelius  introduced  the  sign  —  for  minus;  so  that 
20 — 6,  is  read  20  minus  6 ;  and  signifies,  20  with  6  sub- 
traded  from  it.     Thus,  20—6=14. 

Example  7.  At  a  certain  election,  548  persons  voted  ; 
and  the  successful  candidate  had  a  majority  of  130  votes. 
How  many  voted  for  each  ? 

Let  us  represent  the  number  of  votes  received  by  the  suc- 
cessful candidate,  by  the  letter  A.  Then,  as  the  unsuccess- 
ful candidate  received  1,30  votes  less,  his  number  may  be 
represented  by  Ji — 130. 

Then  A,+A— 130,=548 

Putting  the  A's  together,  2 A— 130=548  which 

means,  2 A,  after  130  is  subtracted  from  it,  is  equal  to  548. 
Then,  2A,  before  130  was  subtracted  from  it,  was  equal  to 
130  more  than  it  is  now.  That  is,  it  was  equal  to  548  and 
130  more.     Therefore  we  say  adding  the  130,     2A=678. 

Dividing  by  2,       A=339. 
The  successful  candidate  had         339. 
The  unsuccessful  one,  130  less=209. 

Proof  548. 
§9.  Des  Cartes*  a  Frenchman,  who  wrote  about  1637, 
used  the  last  letters  of  the  alphabet ;  namely,  x,  y,  z,  u,  &c. 
to  denote  the  unknmvn  quantities.     And  this  is  now  the 
practice  of  all  mathematicians. 

•  Pronounced  Da-Cart'. 

2* 


18  ALGEBRA. 

f 

Example  8.  It  is  required  to  divide  $300  among  A,  B, 
and  C;  so  that  *^  may  have  twice  as  much  as  B,  and  C 
may  have  as  much  as  *i  and  B  together. 

Let  us  represent  B's  share  by  x.  Then  A's  share  will 
be  2x;  and  C  will  have  as  much  as  both  put  together,  which 
is  Sx.     Then, 

a:+2ar+3ar=300. 
Putting  the  x's  together,  6ar=300. 

Dividing  by  6,  x=  50. 

Therefore,  B's  share  is         $50. 
A's  is  twice  as  much,  or        8100. 
C's=.as  much  as  both,  or     $150. 


Proof,  $300. 
§10.  As  we  sometimes  wish  to  speak  of  particular  parts  of 
our  calculations,  mathematicians  have  given  the  name  term 
to  any  quantity  that  is  separated  from  others  by  one  of  the 
SIGNS  +  or  — .  Thus,  in  the  last  Example,  and  first  line 
of  the  operation,  the  first  x  is  the  first  term,  the  2x  is  the 
next  term,  and  the  Sx  is  the  next  term ;  and  the  300  is  the 
last  term. 

§11.  When  a  figure  is  put  before  a  letter  to  denote  how 
many  times  we  take  the  quantity  which  that  letter  stands 
for,  the  figure  is  called  a  co-ejlcient.^  Thus,  in  the  term 
2ar,  2  is  a  co-efficient  of  x ;  in  the  term  2^,  3  is  a  co-efli- 
cient  of  d.  ^ 

§12.  It  must  also  be  understood,  that  a  letter  without  any 
number  before  it,  has  1  for  its  co-efficient.  Thus,  x  re- 
presents la:;  a=la;  &c.  The  1  is  omitted  because  it  is 
plainly  to  be  understood. 

•  This  name  was  given  by  Franciscus  Vieta,  about  1573. 


ADDITION  AND  SUBTRACTION.  19 

§13.  Any  number,  or  letter,  or  any  thing  else,  used  to 
denote  a  quantity,  when  it  is  not  united  to  any  other  quan- 
tity by  either  the  sign  +  or  — ,  is  called  a  simple  quantity. 
Thus,  a;  is  a.  simple  quantity ;  2x  is  a  simple  quantity  ;  12 
is  a  simple  quantity  ;  &c. 


ADDITION   AND    SUBTRACTION  OF  SIMPLE    QUANTI- 
TIES. 

§14.  In  Algebra,  simple  quantities  are  added  by  writing 
them  down,  one  after  another ;  being  careful  to  put  the 
sign  +  between  them.  Thus  we  add  9  to  x,  by  writing 
them  9-|-a;,  or,  a? +9. 

0:!?"  The  pupil  must  understand  that  x  stands  for  some 
number  ;  but  it  is  often  the  case  that  we  do  not  know  what 
that  number  is. 

§15.  It  will  readily  be  seen,  that  it  is  of  no  consequence 
which  quantity  is  put  first;  for  9 4-7  is  the  same  amount 
as  7  +  9. 

§16.  One  simple  quantity  is  subtracted  from  another 
simple  quantity,  by  writing  down  both  quantities,  one  after 
another,  and  putting  the  sign  —  before  the  quantity  which 
tee  subtract. 

Thus,  we  subtract  4  from  x,  by  writing  them  x  —  4. 

§17.  When  two  simple  quantities  have  been  added,  or 
one  simple  quantity  subtracted  from  another,  they  then  will 
consist  of  more  than  one  term. 

§18.  Quantities  that  consist  of  more  than  one  term,  are 


20  ALGEBRA. 

called  compound  quantities.  Thus,  a-f  &>  is  a  compound 
quantity.     So  is  a — b,  and  ar-}-7,  and  x — 7,  &c. 

§19.  It  sometimes  happens,  that  in  those  compound  quan- 
tities which  are  made  by  adding  or  subtracting,  there  are 
two  or  more  terms  of  the  same  kind  ,*  such  as  x-\'2x, 
4a — 2a,  5x — Sx+x.  Such  quantities  are  called  like  quan- 
tities. 

§20.  When  in  any  compound  quantity,  there  are  two  or 
more  terms  of  the  same  kind,  they  may  be  united  by  per- 
forming the  operation  which  is  expressed  by  the  sign.  Thus, 
x-{-2x  is  united  into  3a:.  4a — 2tt  is  united  into  2a.  5ar — 3a; 
-f-a?  is  united  into  Sx. 

§21.  Numeral  quantities*  may  be  united  in  the  same 
manner.  Thus,  44-3  may  be  united  into  7.  9 — 4  may  be 
united  into  5.  6 — 2-f  5  may  be  united  into  9. 

EXAMPLES. 

Unite,  as  much  as  possible,  the  terms  in  each  of  the  fol- 
lowing compound  quantities. 

1 .  a+  fe + 8—4 + 2a + 6  -f  Sb—b  -f  3a.  Take  one  letter 
and  go  through  the  whole  quantity  with  that  first ;  and  then 
take  another  letter  and  do  the  same.    And  then  another,  &c. 

Ans.  Uniting  the  a's,  they  equal  +  6a ;  uniting  the  &'s, 
they  equal -j- 36  ;  uniting  the  numeral  quantities,  they  equal 
+  10.     Therefore  the  answer  is,  6a  +86  +  10. 

2.  2a+6+4a+26+8+6— 3a— 6— 4. 
8.  Sx-}-y+z+3x-{-Sy  +  z-i-^x—2y-]-z. 

4.  18— 9+5+8a+4a:+7a;+5a— 6a:— 7a. 

•  Numeral  quantities  are  expressed  hy  figures  i  and  literal 
quantities,  by  letters. 


ADDITION  AND  SUBTRACTION.  21 

5.  4y4-72j  +  9a— 5a— 42J— y4-2s+3a4-2y. 

6.  la — a — a+5&+4«+a — 2z+a — z-\-Aa, 

7.  8— 4+6+5— 24-8+a;4-3a7— 2a?+4a?. 

§22.  When  any  simple  quantity  begins,  with  the  sign  +> 
it  is  called  a  positive  quantity  ;  as  +a,+3a. 

§23.  When  any  simple  quantity  begins  with  the  sign  — , 
it  is  called  a  negative  quantity  ;  as,  — 6,  — 5a. 

§24.  In  algebra,  the  perfect  representation  of  any  sim- 
ple quantity  requires  both  the  specified  sum,  and  either  the 
sign  +,  or  the  sign  — ;  as,  +5,  — 40,  +ar,  — 3ar. 

§25.  But,  when  a  positive  quantity  stands  by  itself,  or 
when  it  is  the  first  term  of  a  compound  quantity ;  the  sign 
that  belongs  to  it,  is  generally  omitted  on  paper,  and  also 
in  our  reading;  as,  x,    2,  a+&,    x — y. 

§26.  Therefore,  when  a  simple  quantity,  or  the  first  term 
of  a  compound  quantity,  does  not  begin  with  a  sign,  we 
say  that  the  sign  +  is  understood.  That  is,  we  think  of 
the  quantity  the  same  as  if  +  was  before  it. 

§27.  In  reading  compound  quantities,  the  pupil  must  be 
careful  to  join  the  sign  to  the  term  that  is  immediately  after 
it.  Thus,  the  first  example  under  §21,  must  be  read  a  ; 
plus  b  ;  plus  8  ;  minus  4  ;  plus  2a  ;  &c. 

We  are  now  ready  for  the  following 

GENERAL  RULE  FOR  UNITING  TERMS. 

§28.  Select  one  kind  of  like  quantites,  and  add  into  one 
sum  all  the  positive  co-efficients  that  belong  to  them.  Then 
add  into  another  sum  all  the  negative  co-efficients  that  belong 
to  them.  Then  subtract  the  less  sum  from  the  greater; 
and  prefix  the  sign  of  the  greater  lo  the  difference,  annexing 
the  common  letter. 


22  ALGEBRA. 

Note. — It  sometimes  happens  that  the  negative  quantity  is 
greater  than  the  positive  quantity.  In  such  cases,  the  difference 
will  have  the  sign  — . 

Examples.     Unite  the  terms  in  the  following. 

1.  3a  +  46+26— 3c+3c— &+a. 
Ans.  1st,      4a+6&— &  +  3c~3c. 

Now  +66 — &=  +  56;  and  +3c — 3c  balance  each 
other,  so  as  to  be  equal  to  0.     The  final  answer  is  4a +56. 

2.  a+6+3a— c+4a — 3c+6.  Ans.  8a+26— 4c. 

3.  5a? +51/  + 3a;— 2y.  Ans.  8a: +3^. 

4.  6+a+3a — 56+ 4a; — a  +  3a — x.   Ans.  6a — 46+ 3a;. 

5.  4a— 56— 10  +  2?/— a;+4.   Ans.  4a— 56+22/— a;— 6. 

6.  2x-\-Sy — 3a?+53/ — x — z.  Ans.  — 2x-^8y — z. 

7.  x-\-z+x — z-\-x-\-2y.  Ans.  3a;+3y. 

8.  3a— 5— 46  +  6— 2a— 3  +  66.  Ans.  a +26— 2. 

9.  4a;+4+5i/+6— 2  +  3a;— 21/.  Ans.  7a? +32/ +8. 
^    10.  4c+3a — 6 — 3a+c — 2a— 6.        Ans.  5c— 2a — 26. 

11.  26— c+3a— 6+4c+2<i— 6— 5t?. 

12.  a— 6+c+(^+3a+6— 2c— 4a+10+a. 

13.  a  +  6+36+a?— 7+4a;— 6  +  3a— 2/+2. 

14.  ar— 4+10— 7a;+a— 6— 10+2a— 1. 

15.  3a— 6+4c+2a;— c+4a  +  5c+76. 

16.  10— a+6— 6— a?+7  +  36+2a— 40. 

17.  8a— 16— z  +  91+2y— 87— 32  +  14. 

18.  29+46+27+2/— 32— 43+3/. 

19.  73+4a?— 36— .3aj— 41+7a;— 3/+2a;, 

20.  a--'6— a--6— 67— 42+7a  +  36. 


MULTIPLICATION  AND  DIVISION.  23 


MULTIPLICATION   AND    DIVISION  OF  SIMPLE   QUAN- 
TITIES. 

§29.  Were  we  to  add  a  to  itself  four  times,  we  should 
write  the  sum  thus,  a-^a-\-a-\-a;  which,  when  united,  be- 
comes 4a.  Whence  we  see  that  any  literal  quantity  is 
multiplied  by  a  number,  by  putting  that  number  before  it  as 
a  co-efficient ;  as,  7  times  x  is  Ix.  6  times  a  is  Qa. 

The  pupil  may  now  multiply  x  by  each  of  the  numbers 
from  2  to  20. 

§30.  If  the  quantity  to  be  multiplied  has  already  a  co- 
efficient, that  co-efficient  only  is  to  be  multiplied.  Thus,  3a; 
taken  four  times  is  Sx-i-3x-\-Sx-\-SXj  which  when  united 
=12a:.  The  co-efficient  is  multiplied  by  4.  Thus,  4  times 
3a:=12a:. 

The  pupil  may  multiply  2x  by  each  number  frora  2  to  12. 

He  may  then  multiply  3a:  by  the  same  numbers  ;  and 
then  4a:  by  the  same. 

§31.  It  is  evident  that  if  2  times  3a?  is  6a?,  then  one  half 
of  6a:  is  3a:.  Whence  we  learn  that  a  quantity  with  a  nu- 
meral co-efficient,  may  be  divided,  by  merely  dividing  that 
co-efficient.  Thus,  8a?  divided  by  2=4a?.  12a:  divided  by 
4= 3a:,  &c. 

§32.  In  1661,  Rev.  William  Oughtred  of  England,  pub- 
lished a  work,  in  which  he  introduced  the  sign  X  to  repre- 
sent multiplication.  Thus,  4x3=12,  is  i^^A  4c  multiplied 
by  3  equals  12  ;  or,  4  into  3  equals  12. 

§33.  In  1668,  Mr.  Brancker  invented  the  sign  —  for  di- 
vision. This  sign  is  always  put  before  the  divisor;  as 
20-j-4=5;  read  20  divided  by  4  equals  5  ;  or,  20  hy  4, 
equals  5. 


24  '  ALGEBBA. 

SIMPLE  EQUATIONS. 

§34.  The  most  general  application  of  algebra,  is  that 
which  investigates  the  values  of  unknown  quantities  by 
means  of  c^wa^iows.  'n  v 

§35.  An  equation  is  an  expression  which  declares  one 
quantity  to  be  equal  to  another  quantity,  by  means  of  the 
sign  ==  being  placed  between  them. 

Thus,  6+3=8,  is  an  equation,  denoting  that  5  with  3 
added  to  it,  equals  8.  Also,  4 — 1=3,  and  3+2 — 1=4, 
and  8 — 2=5  +  1,  are  equations,  each  denoting  that  the 
quantity  on  one  side  of  the  =,  is  equal  to  that  on  the  other 
side. 

§36.  The  whole  quantity  on  the  left  of  =  is  called  the 
first  member  of  the  equation  ;  and  all  on  the  right  of  =  is 
called  the  last  member  of  the  equation. 

In  order  to  be  a  member  of  the  equation,  it  is  of  no  im- 
portance whether  the  quantity  is  simple  or  compound.  Thus, 
in  the  equation,  a?=4+a — b — 18,  x  is  the  first  member, 
and  4+a — b — 18  is  the  last  member.  And  in  this  case, 
the  first  member  represents  just  as  great  a  quantity  as  the 
last. 

§37.  In  order  that  an  equation  may  be  such  that  we  can 
find  the  value  of  an  unknown  quantity  by  it,  it  must  con- 
tain some  quantity  that  is  already  known.  And  then,  we 
find  the  value  of  the  unknown  quantity,  by  making  that 
stand  by  itself  on  one  side  of  =,  and  all  the  known  quan- 
tities on  the  other  side  ;  taking  care  to  change  them  in  such 
a  manner  as  not  to  destroy  the  equation. 

§38.  The  operation  of  managing  an  equation,  so  as  to 
bring  the  unknown  quantity  to  stand  equal  to  a  known 
quantity,  is  called  solving  or  reducing  the  equation. 


SIMPLE  EQUATIONS.  25 


EQUATIONS. — SECTION  1. 

Equations  which  are  solved  by  merely  uniting  terms. 

In  each  of  the  following  equations,  the  object  is  to  find 
the  value  of  x. 

Example  1.     a;+2a7=45— 15. 
Uniting  terms,  3iC=30. 

Now,  as  we  have  found  that  three  a?'s=30 ,  it  is  evident 
that  one  x  will  be  one  third  of  30.  Therefore,  dividing  by 
3,  ir=10. 

2.  8a:— 4a;— a;=7+26+51— -15 

Uniting  terms,  3a;=69 
Dividing  by  3,    ar=23. 

3.  10a:— 5a;+4a;=56+75  +  32— 1 

Uniting  terms,  9a:=162 
Dividing  by  9,     a:=18. 

4.  a:+2a:+3a:+4a:=12+35+74— 11 

Uniting  terms,  10a:=nO 
Dividing  by  10,    a:=ll. 

5.  8a:— 3a:+2a:=46+54  +  37— 4.  Ans.  a:=19. 

6.  4a:— 3a:+4a:=29— 36+48+14.  Ans.  a:=ll. 

7.  6a-— 8a:+14a:=12+ 36+14+22.  Ans.  a:=7. 

8.  .5a:+4a:+3a:=49+14+22  +  ll.  Ans.  a:=8. 

9.  7a:+a:=14— 22— 11+41— 6.  Ans.  a:=2. 

10.  4a:— 2a:=96— 7+8— 15— 10.  Ans.  a:=36. 

11.  5a:— a:=2+3— 15— 10+72.  Ans.a:=13. 

12.  6a:=7+4+72— 51— 16— 10.  Ans.a:=:i. 

13.  8a:— 7a:+5a:— 4a:+3a:=27— 12.  Ans.a?=3. 

14.  5a:— 4a:+2a:— 3a;+a:=39— 13.  Ans.  a: =26. 

3 


26  ALGEBRA. 

15.  17^7 — 4a? — 3a; — 5a; — a;=57 — 32.  Ans.  a;=5# 

16.  14a;— 36a;+29a;+47a;+a;=504*  Ans.a;=9. 

PROBLEMS. 

§39.  An  algebraic  problem  is  a  proposition  which  requires 
the  discovery  or  demonstration  of  some  unknown  truth. 

§40.  In  the  solution  of  problems,  the  first  thing  to  be 
done,  is  to  make  a  statement  of  the  conditions,  in  algebraic 
language,  in  the  same  manner  as  if  the  answer  were  al- 
ready found,  and  you  were  required  to  see  if  it  is  right.  In 
order  to  do  this,  it  is  customary  to  represent  the  unknown 
quantity  by  a:,  y,  or  some  other  final  letters  of  the  alphabet. 

§41.  When  the  question  has  been  fairly  stated,  it  will  be 
found  that  some  condition  has  been  represented  in  two  ways ; 
one  having  the  unknown  quantity  in  it,  and  the  other  hav- 
ing a  known  quantity.  These  two  expressions  must  be  put 
together,  so  as  to  form  an  equation.  And  then,  by  reducing 
the  equation,  the  required  result  will  be  found. 

1.  The  sum  of  $660  was  subscribed  for  a  certain  pur- 
pose, by  two  persons,  A  and  B ;  of  which,  B  gave  twice 
as  much  as  A.     What  did  each  of  them  subscribe  ? 

Stating  the  question,  A  gave  x  dollars. 

B  gave  2a;  dollars. 
Both  of  them  gave  x-{-2x  dollars. 

•  But  both  gave  660  dollars. 
Therefore,  putting  the  question  into  an  equation, 

a;+2a;=660 
Uniting  terms,  3a;==660 

•  Dividing  by  3,    a;=220  A's  share. 

2a;=440  B's  share. 

660  proof. 

2.  Three   persons  in  partnership,    put  into  the  stock 


SIMPLE  EaUATIONS.  27 

^4800;  of  which,  A  put  in  a  certain  sum,  B  twice  as 
much,  and  C  as  much  as  A  and  B  both.  What  did  each 
man  put  in  1 

Stating  the  question,  ^==A's  share. 

2a?=B's  share. 

a?+2a?=C's  share. 

x-i-2x-\-x-\-2x—  the  whole. 

4800=  the  whole. 

Therefore,  forming  the  equation, 

x+2x-i'X+2x=-AS00 
Uniting  terms,  6a?=4800 

Dividing  by  6,  rp=800    A's  share. 

2ar=1600  B's  share. 
800  +  1600=2400  C's  share. 

4800  proof. 
Ot!F  In  all  the  succeeding  problems,  the  learner  should 
prove  his  answers. 

3.  A  person  told  his  friend  that  he  gave  108  dollars  for 
his  horse  and  saddle ;  and  that  the  horse  cost  8  times  as 
much  as  the  saddle.     What  was  the  cost  of  each  ? 

Stating  the  question,  x=:  price  of  the  saddle. 

8a;=         '*  horse. 

x-{-8x=z         "  both. 

108=  «  do. 

Forming  the  equation,  a;+8a;=108 
Uniting  terms,  9a7=108 

Dividing  by  9,  a?=12=  price  of  saddle. 

It  is  advisable  for  the  pupil  while  performing  his  sums,  to 
write  them  on  his  slate  in  a  manner  similar  to  the  three 
questions  above ;  beginning  the  statement  by  making  x  the 
answer  to  the  question.  And  in  recitation,  the  whole  of  it 
is  to  be  recited. 

4.  A  father  once  said,  that  his  age  was  six  times  that  of 


ALGEBRA. 


his  son ;  and  that  both  of  their  ages  put  together,  would 
amount  to  49  years.  What  was  the  age  of  each?  Ans. 
Son's  age  7  years ;  father's  42. 

5.  A  farmer  said  that  he  had  four  times  as  many  Cows  as 
horses,  and  five  times  as  many  sheep  as  cows ;  and  that 
the  number  of  all  of  them  was  100.  How  many  had  he 
of  each  sort.     Ans.  4  horses;  16  cows  ,*  and  80  sheep. 

6.  A  boy  told  his  sister  that  he  had  ten  times  as  many 
chestnuts  as  apples,  and  six  times  as  many  walnuts  as 
chestnuts.  How  many  had  he  of  each  sort,  supposing 
there  were  639  in  all.  Ans.  9  apples;  90  chestnuts;  and 
540  walnuts. 

7.  A  school  girl  said  that  she  had  120  pins  and  needles; 
and  that  she  had  7  times  as  many  pins  as  needles.  How 
many  had  she  of  each  sort?  Ans.  15  needles,  and  105 
pins. 

8.  A  teacher  said  that  her  school  consisted  of  64  scholars ; 
and  that  there  were  three  times  as  many  in  arithmetic  as  in 
algebra,  and  four  times  as  many  in  grammar  as  in  arith- 
metic. How  many  were  there  in  each  study  ?  Ans.  4  in 
algebra;  12  in  arithmetic;  and  48  in  grammar. 

9.  Two  men;  who  are  560  miles  apart,  start  to  meet 
each  other.  One  goes  30,  and  the  other  goes  40  miles  a 
day.     In  how  many  days  will  they  meet  ?     Ans.  8  days. 

Remarks,  Each  will  travel  x  days.  The  first  will  go 
X  times  30  miles,  and  the  second  will  go  x  times  40  miles  ; 
and  both  together  will  go  the  whole  distance.  It  is  also 
evident  that  x  times  30  is  the  same  as  30  times  x ;  &c. 

10.  A  teacher  had  four  arithmeticians  who  performed  80 
sums  in  a  day.  The  second  did  as  many  as  the  first,  the 
third  twice  as  many,  and  the  fourth  as  much  as  all  the 


SIMPLE  EatTATIONS.  29 

Other  three.     How  many  did  each  perform?     Ans.  The 
first  and  second,  each  10  ;  the  third,  20 ;  and  the  fourth,  40 

11.  A  person  said  that  he  was  $450  in  debt.  That  he 
owed  A  a  certain  sum,  B  twice  as  much,  and  C  twice  as 
much  as  to  A  and  B.  How  much  did  he  owe  each  ?  Ans. 
To  A  $50,  to  B  8100 ;  and  to  C  8300. 

12.  A  person  said  that  he  was  owing  to  A  a  certain 
sum  ;  to  B  four  times  as  much  ;  and  to  C  eight  times  as 
much  ,*  and  to  D  six  times  as  much  ;  so  that  8570  dollars 
would  make  him  even  with  the  world.  What  was  his  debt 
to  A?     Ans.  830. 

13.  A  man  bought  3  sheep  and  2  cows  for  860.  For 
each  cow,  he  gave  6  times  as  much  as  for  a  sheep.  How 
much  did"  he  give  for  each  ? 

Remarks,  l^  x=  price  of  a  sheep,  all  the  sheep  will 
cost  three  times  as  much,  or  3a:.  In  the  same  manner 
both  cows  will  cost  twice  as  much  as  one  cow.  One  cow 
will  cost  6x,  and  2  cows  will  cost  12a;.  Ans.  84,  price 
of  a  sheep ;  and  824  price  of  a  cow. 

14.  A  gentleman  hired  3  men  and  2  boys  for  one  day. 
He  gave  five  times  as  much  to  a  man  as  he  gave  to  a  boy ; 
and  for  all  of  them  he  gave  86.80.  What  was  the  wages 
of  each  1  Ans.  A  boy's  wages  was  40  cents,  and  a  man's 
wages,  82. 

15.  A  boy  bought  some  oranges  and  some  lemons  for  54 
cents.  There  was  an  equal  number  of  each  sort,  but  the 
price  of  an  orange  was  twice  the  price  of  a  lemon.  How 
much  money  did  he  spend  for  each  sort?  Ans.  18  cents 
for  lemons ;  and  36  cents  for  oranges. 

16.  A  boy  bought  some  apples,  some  pears,  and  some 
peaches,  an  equal  number  of  each  sort,  for  72  cents.  The 
price  of  a   pear  was  twice  that  of  an  apple,  and  the  price 

3* 


810  '^^    ..ij.yU'GEBRA. 

of  a  peach  was  3  times  that  of  an  apple.  How  much  mo- 
ney did  he  give  for  each  kind?  Ans.  12  cents  for  apples; 
24  cents  for  pears  ;  36  cents  for  peaches. 

17.  A  farmer  hired  three  labourers  for  $50.00 ;  giving 
lo  the  first  82.00  a-day,  to  the  second  $1.50,  and  to  the 
third  $1.00.  The  second  worked  three  times  as  many  days 
as  the  first ;  and  the  third  twice  as  many  days  as  the  se- 
cond. How  many  days  did  each  work  ?  Ans.  The  first, 
4;  second,  12  ;  and  third,  24  days. 

18.  A  gentleman  bought  some  tea,  coffee,  and  sugar,  for 
$7.04  ;  giving  twice  as  much  a  pound  for  coffee  as  for  sugar, 
and  five  times  as  much  for  tea  as  forcofl^ee  ;  and  there  were 
20  pounds  of  sugar,  12  pounds  of  coffee,  and  3  pounds  of 
tea.  What  was  the  price  of  each?  Ans.  11  cents  for 
sugar;  22  cents  for  coffee;  and  110  cents  for  tea. 


ADDITION  OF  COMPOUND  QUANTITIES. 

§42.  When  two  or  more  expressions  that  consist  of  seve- 
ral terms,  are  to  be  added  together,  the  operation  is  repre- 
sented by  connecting  them  with  one  another  by  means  of 
the  sign  -f .     Thus,  a — x  is  added  to  y-{-7  in  the  following 

manner : 

a — x-i-y+l,  or  y+7+a — x. 

For  to  a — x  we  add  first  y,  and  then  7  ;  or  to  y+7  we  add 
a,  and  then  because  we  ought  to  have  added  a — ar,  we  see 
that  we  have  added  x  too  much,  and  therefore  subtract  it. 

§43.  The  above  example  shows  that  it  is  of  no  conse- 
quence in  what  order  we  write  the  terms.  Their  place  may 
be  changed  at  pleasure,  provided  their  signs  be  preserved. 

§44.  After  compound  quantities  have  been  added,  their 
terms  may  be  united  according  to  the  rule  §28. 


ADDITION  OF  COMPOUND  QUANTITIES.  31 


EXAMPLES. 

1.  Add  the  following  compound   quantities.     2a — 8x, 
X — 3a,  — ia — 2x,    4a? — a.     Ans.  — 6a — 5ic. 

2.  Add  2— a:+4y,     3+3ar— ^,      — 30— a>— 2y,    and 
1— 2a:-f  Sy—lOa;  together.     Ans.  24— a;-f  4y-— IO2;. 

3.  Add   3ar+5i/— 6z,     — 2a:— 8y— 9z,     20ir+2i/— 3z 
and  X — y-j-z — 4  together.     Ans.  22a; — 2y — 173 — 4. 

4.  Add  3—23/4-2,    4y— 2z+5,     2—2—3/,   ^"^  2z— y 
— 10.     Ans.  Nothing. 

5.  Add  7a? — 61/ — 5z — 8—^ 

S—g—Sy—x 

7^— 1— 32-fi/— a; 

32— 1—^-1-31/— 2a; 

a?+82/— 52+9+^.     Ans.  4a;+5^4-3y+2. 

6.  Add  Zb—a—c—ll^d+ee—dy 

Za-\-21e—d—Zc~-2h 
3e— 7i/— 8c+56 
17c— 6&— 7a+9rf— 5c+lly 
— 2«Z — 6e— 5c — 93/— 3a+ar. 

Ans.  37e— 8a— 109(f— 101/4- X. 

7.  Add  4— 3a4-a?— 7443/ 

84-63/— 3a:46a432 

7a433/4-4a4-12— a? 

X — 4483/+6a4n  '      ^ 

4a  4  7  4- 3a  4  iP— 23/    Ans. 

8.  Add  a — x-\-y-\-r — 14 — n 

3n— 3/4246a— 2+3a:+r 

6247r — a—y—2n 

3y42a4146r44» 

a?— 74^4^44 — «4r.    Ans. 


^.. 


42  -Ravim?  ALGEBRA. 

TRANSPOSITION. 

1.    BY  SUBTRACTION. 

§45.  It  often  happens  that  in  the  first  member  of  the 
equation,  some  number  has  been  added  to  the  x's  in  order 
to  make  them  equal  to  the  last  member.  Thus,  in  the 
equation,  ar-J-16=46,  we  see  that  16  has  been  added  to  .r, 
to  make  it  equal  to  46. 

§46.  Now  if  X  with  1 6  added,  is  equal  to  46  ;  then  x  alone 
must  be  16  less  than  46 ;  that  is,  46 — 16.  So,  that  if  we 
find,  that  a: -f- 16=46,  we  may  know  that  a:=46 — 16;  or 
what  is  the  same,  a:=30. 

§47.  But  this  may  be  proved  another  way.  It  is  very 
plain  that  if  we  subtract  a  quantity  from  one  member  of  an 
equation,  and  then  subtract  the  same  quantity  from  the 
other  member  of  the  equation;  it  will  still  be  the  fact  that 
the  two  members  are  equal  to  one  another.  Thus,  a  half 
dollar=  50  cents.  Subtract  2  cents  from  each  member. 
Then  a  half  dollar  —  2  cents  =  50  cents  —  2  cents ;  for 
each  of  them  is  equal  to  48  cents. 

§48.  Now,  with  the  equation  that  we  had  above, 

ar+16=46. 
Subtracting  16  from  both  members,   x-\-lQ — 16=46 — 16. 

Now,  in  the  first  member  of  the  equation,  we  have+16 
— 16,  which  is  of  no  value  at  all,  for  +16  and  — 16 
balance  each  other  as  has  been  seen  in  Ex.  1  under  §28. 
Therefore  the  equation  is  reduced  to  a; =46 — 16. 

Uniting  terms  in  the  last  member,  ar=30. 

EQtJATIONSi-^SECTION  2. 

1.  Suppose  a'+8  +  3x-»56  ;  what  is  the  value  of  a;? 
Uniting  terms,  4a? 4- 8 =56. 


iW 


TRANSPOSITION  BY  SUBTRACTION.  33 

Subtracting  8  from  both,         4a;4-8— 8=56— 8. 
Which  is  the  same  as  4a;=56 — 8. 

Uniting  terms  in  the  last  member,    4a?=48. 
Dividing  by  4,  a;=12. 

2.  Suppose  2a:+14— a;— 7=41+2— 8  ;  to  find  x. 

Uniting  terms,     a;-f7=35. 

Subtracting  7  from  i      x-\-l — 7=35 — 7  or 

both  sides.       \      a:=35 — 7. 
Uniting  terms,  ic=28. 

3.  Given  a; +5 =6,  to  find  x. 

Writing  the  equaiion,  a;-f  5=6. 
Subtracting  5,  a?=l. 

4.  Given  .5a^+22— 2a:=31,  to  find  x.  Am,  x=S, 

5.  Given  4a?+20— 6=34,  to  find  x.  Ans.  x=5. 

6.  Given  3a?+12+7a;=102,  to  find  x.        Ans.  x=9, 

7.  Given  10a;— 6a:+14=62,  to  find  x,        Ans,  x=\2, 

8.  If7a;— 14+5a;+20=246,  thena?=20. 

9.  If8a?4-17— 5ar+3=100+10,  thena?=30. 
10.  If  7a;— 14+3a?+35=450— 29,  then  a;=40. 

PROBLEMS. 

1.  What  number  is  that,  which,  with  5  added  to  it,  will 
be  equal  to  40? 

Staling  the  question,  a;=the  number 

a? +5,=  after  adding. 
Forming  the  equation,  a?4-5=40. 

Subtracting  5,  from  both,  a?=35. 

2.  A  man  being  asked  how  many  shillings  he  had  an- 
swered, add  15  to  their  number,  and  then  subtract  1,  and  the 
remainder  will  be  64.     How  many  shillings  had  he? 

Stating  the  question,  a;=numb.  of  shillings. 

'^'        '  a;+15=after  adding. 

a;+15 — 1= after  subtracting. 


^  ALGEBRA. 

Forming  the  equation,         a?4-15 — 1=64. 
Uniting  terms,  a? +14=  64. 

Subtracting  14  from  both,       a? =50. 

3.  What  number  is  that,  which  with  9  added  to  it,  will 
equal  23?     Ans.  14:.  •^' 

4.  Divide  17  dollars  between  two  persons,  so  that  one 
may  have  4  dollars  more  than  the  other. 

Stating  the  question,  a:=:the  less  share. 

a; +4=: the  greater. 

x+x-{- 4=  both  shares. 
Forming  the  equation,  a:+a;+4=17. 

Uniting  terms,  2a7+4=17. 

Subtracting  4  from  both  sides,  2a?=13. 

Dividing  by  2,  x=6h 

5.  The  sum  of  the  ages  of  a  certain  man  and  his  wife  is 
55  years ;  and  his  age  exceeds  her's  by  7  years.  What  is 
the  age  of  each  ?  0^  Let  a;=age  of  the  wife,  .dns*  24 
the  wife's.     31  the  man's. 

6.  A  is  5  years  older  than  B,  and  B  is  4  years  older 
than  C ;  and  the  sum  of  their  age  is  73  years.  What  is  the 
age  of  each  ? 

Stating  the  question,    ;r=C's  age, 
a?4-4=B's  age. 
a?-f  4+5=:A's  age. 
a:-f  a?4-4+a?+4+5=sum  of  all  of  them. 
Forming  the  equation,         a? -|- a? -f- 4+ a; +4 -{-5=: 7 3. 
Uniting  terms,  3;r+13=73. 

Subtracting  13  from  both  sides,  3a;=60, 

Dividing  by  3,  «;=20. 

Ans,  C  20  years,  B  24,  A  29. 

7.  Two  persons  were  candidates  for  a  certain  office, 
where  there  were  329  voters.     The  successful  candidate 


TRANSPOSITION  BY  ADDITION.  35 

gained  his  election  by  a  majority  of  53.     How  many  voted 
for  each?  Ans.  191  for  one,  and  138  for  the  other. 

8.  A,  B,  and  C,  would  divide  $200  among  themselves, 
so  that  B  may  have  $6  more  than  A  ;  and  C  $8  more  than 
B.  How  much  must  each  have  ?  Ans.  A  must  have  $60, 
B  $76,  and  C  $74. 

9.  Divide  $1000  between  A,  B,  and  C:  so  that  A  shall 
have  $72  more  than  B,  and  C  $100  more  than  A.  Ans. 
Give  B  $252,  A  $324,  and  C  $424. 

10.  At  a  certain  election  1296  persons  voted,  and  the 
successful  candidate  had  a  majority  of  120.  How  many 
voted  for  each  ?       Ans.  588  fo»  one,  and  708  for  the  other. 

11.  A  father,  who  has  three  sons,  leaves  them  $8000, 
specifying  in  his  will  that  the  eldest  shall  have  $1000  more 
than  the  second,  and  that  the  second  shall  have  $500  more 
than  the  youngest.  What  is  the  share  of  each  ?  Ans.  The 
eldest  had  $3500,  the  second  $2500,  the  youngest  $2000. 

12.  A  cask  which  held  74  gallons  was  filled  with  a  mix- 
ture of  brandy,  wine  and  water.  In  it,  there  were  15  gal- 
lons of  wine  more  than  of  brandy,  and  as  much  water  as 
both  wine  and  brandy.  What  quantity  was  there  of  each  ? 
Ans*  11  gallons  of  brandy,  26  of  wine,  and  37  of  water. 


2.    TRANSPOSITION  BY  ADDITION. 

§49.  It  is  frequently  found  that  some  quantity  has  been 
subtracted  from  the  a?'s  ,*  as  in  the  equation     bx — 44=76. 

§50.  In  such  cases,  it  is  very  evident,  that  the  quantity 
which  has  been  subtracted  from  the  a?'s,  must  be  added  to 


35  ALGEBRA. 

each  side.     For  if  in  the  above  equation,  44  has  already 
been  subtracted  from  the  5x ;  we  must  add  it  again  if  we 
wish  to  find  what  bx  is  equal  to.     But  if  we  add  44  to  one 
member  of  the  equation,  we  must  also  add  as  much  to  the 
other  member  of  the  equation.     So  that  it  will  become 
5a;— 44+44=76+44. 
Uniting  the  two  44's,        5a;=76-f  44. 
Or,  5a?=120. 

A        Dividing  by  5,  ic=24. 

iJ!^  EQUATIONS. SECTION  3. 

1.  Given  a?+ 14+ 3a;— 27=51,  to  find  x.  Ans,  a;=16. 

JV  Given-3a?— 30— 2ar=:46*-7,  to  find  x.  Ans,  ar=:69. 

3.  Given  9a;—  41  +6=88—6,  to  find  x,  Ans.  a;=l3. 

4.  Given  20  +  3a;— 46=35— 4,  to  find  x.  Ans.  a;=l9. 

5.  Given  4a;— 39 — 2a; =47,  to  find  x.  Ans.  a;=43. 

6.  Given  7a;+27— 46=65,  to  find  x.  Ans.  a;=12. 

7.  Given  14a;— 55— 8a; +14=  85,  to  find  x,  Ans,  a;=21. 

PROBLEMS. 

1.  What  number  is  that,  from  which  8  being  subtracted, 
the  remainder  is  45  ? 

Stating  the  question  x=  the  number. 

X — ^"8=:  when  8  is  subtracted. 
Forming  the  equation  x — 8=:45 
Adding  8  to  both  sides      a;— 8+8=45+8. 
Or,        a;=53. 

2.  What  number  is  that,  from  which  27  being  subtracted, 
the  remainder  is  41 1  Ans.  68. 

3.  A  person  bought  two  geese  for  $1.40 ;  and  gave  16 
cents  more  for  one  than  he  did  for  the  other.  What  did 
each  cost  him  ? 


J--*' 


TRANSPOSITION  BY  ADDITION.  87 

0^  In  this  and  the  following  questions  of  this  section,  x 
must  stand  for  the  greatest  quantity. 

Stating  the  question,  x=  the  dearest. 

X — 16=  the  cheapest. 
x-\-x — 16=  cost  of  both. 
Forming  the  equation,  x-\-x — 16=140 

Uniting  terms,  2x — 16=140 

Adding  16  to  both  sides,  2ar=1404-16 

Uniting  terms,  2ie=156 

Dividing  by  2,  a?=78 

Ans.  78  cents,  and  62  cents. 

4.  Three  men,  who  are  engaged  in  trade,  put  in  $2600 
as  follows:  A  put  in  a  certain  sum  ;  B,  $60  less  than  A  ; 
and  C,  as  much  as  A  and  B,  lacking  $100.  What  was 
each  man's  share  ? 

Ans.  A's  $705  ;  B's  $645  ;  C's  $1250. 

5.  A  purse  of  $8000  is  to  be  divided  among  A,  B,  and 
C  ;  so  that  B  may  receive  $276  less  than  A,  and  C  $1112 
less  than  A  and  B  together.     What  is  each  man's  share  ? 

Ans.  A's  $2416;  B's$2140;  C's  $3444. 

6.  A  father  has  willed  to  his  four  sons  $25200  as  follows. 
To  D  a  certain  sum  ;  to  C  as  much  as  to  D,  lacking  $550  ; 
to  B  as  much  as  to  C,  together  with  $1550 ;  and  to  A  twice 
as  much  as  to  B,  lacking  $10000.  How  much  does  each 
of  them  receive  ? 

Ans.  A  $5100;  B  $7550  ;  C$6000;  D  $6550. 

7.  Divide  the  number  60  into  three  such  parts,  that  the 
first  may  exceed  the  second  by  8,  and  the  third  by  16. 

Ans.  12;  20;  and  28. 

8.  Three  men  having  found  a  purse  of  $160,  quarreled 
about  the  distribution  of  it.     Atler  the  quarrel,  it  was  found 

4 


38  ALGEBRA.  [Eq.  SeC.  X 

that  A  had  got  a  certain  sum,  and  that  B  had  $30  more 
than  A,  but  C  850  less  than  A.     How  much  did  each  obtain? 

Ans.  A  $60  ;  B  $90 ;  C  $10. 


III.     TRANSPOSITION  OP  THE  UNKNOWN  QUANTITY. 

§51.  We  have  found  that  when  any  term  has  the  sign  + 
it  may  be  removed  from  one  member  of  the  equation  to  the 
other,  if  we  take  care  to  change  the  sign  to  — ;  for  this 
has  been  done  every  time  we  have  subtracted  a  term  from 
both  sides. 

Th«s,  in  the  equation  [x-\-5=i20  ;]  if  we  subtract  5  from 
both  sides,  it  is  plain  that  the  first  member  becomes  x,  and 
the  last  member  becomes  20 — 5  ;  so  that  the  equation 
would  become 

ir=20— 5. 

§52.  So  also  any  term  that  has  the  sign  —  may  be  re- 
moved from  one  member  to  the  other,  if  we  take  care  to 
change  the  sign  to  4-»  Because  this  is  the  same  as  adding 
that  term  to  both  sides.  • ,  "  :  ,    ; 

Thus,  in  the  equation  x — 5=20, 

if  we  add  5  to  both  sides,  the  first  member  becomes  x,  and 

the  last  member  becomes  20-}-5.     So  that  the  equation 

becomes 

a?=20+5. 

§53.  When  we  remove  a  term  from  one  member  of  an 
equation  to  the  other  member,  we  say  that  we  transpose  that 
term  ;  and  the  operation  of  doing  it,  is  called  transposition. 

§54.  It  was  stated  in  §32,  that  an  equation  must  be 
brought  so  that  the  unknown  quantity  will  occupy  one  mem- 
ber of  the  equation,  and  the  known  quantities  embrace  the 
other  member.      And,  as  it  frequently  happens  that  the 


TRANSPOSITION  BY  ADDITION.  39 

unknown  quantities  are  on  both  sides,  we  are  obliged  to  re- 
sort to  transposition  in  order  to  make  one  side  free  from 
them.  And  likewise,  it  is  often  necessary  to  transpose 
known  quantities  from  the  member  which  contains  the  un- 
known quantity. 

§55.  Any  term  may  be  transposed  from  one  member  of 
an  equation  to  the  other,  care  being  taken  to  change  the 
sign  when  we  change  the  side* 

EQUATIONS. SECTION  4. 

1.  Reduce  the  equation  4a; — 14=3a?+12 
Solution.     Transposing  3a?,     4a; — 3a; — 14=12 

Transposing  14,     4a; — 3a;=12  +  14 

Uniting  terms,  a;=26. 
§56.  In  transposing,  it  is  generally  best  to  write  first  the 
unknown  quantity  that  is  on  the  left ;  and  then  bring  over 
those  which  are  on  the  right,  if  there  are  any  there.  Then 
write  those  known  quantities  that  are  already  in  the  right 
hand  member,  and  then  transpose  after  them  what  known 
quantities  there  are  in  the  left. 

2.  Given  21 — 7a;:=40 — llx,  to  find  x,     Ans.  a?=4|. 

3.  Given  40— 6z=136— 14z,  to  find  z.      Ans.  z=l2. 

4.  Given  t/4-12=3i/ — 4,  to  find  y.  Ans.  y=8. 

5.  Given  5a; — 15=2a;4-6,  to  find  x,         Ans.  x=7, 

6.  Given  40— 6a;— 16=120— 14a^,  to  find  x. 

Ans.  x=  12. 

7.  Given  4 — 9i/=14 — lly,  to  find  y.         Ans.  y==5. 

8.  Given  a;-f-18=3a? — 5,  to  find  x. 

Solution.     Transposing  3a;,  x — 3a; +18=  — 5 

Transposing  18,  x — 3a;=  — 5 — 18 
Uniting  terms,  — 2a?  = — 23 

Dividing  by  2,  — a;=  — lid* 


40  ALGEBRA.  [Eq.  SeC.  4. 

§57.  It  is  of  no  consequence  what  sign  accompanies  the 
final  result ;  as  the  magnitude  of  the  quantity  is  not  affected 
by  the  sign.  If  we  remember  that  +  is  understood  and 
may  be  written  with  every  positive  quantity,  it  will  be  very 
evident  that  the  equation  — a:  =  — 1\^  is  just  as  good  as  the 
equation  -fiC=H-ll-|.  In  both  cases,  the  quantity^?  is 
equal  to  the  number  11|. 

§58.  In  the  result  of  the  last  question,  11|^  maybe  trans- 
posed to  the  first  member ;  and  x  may  be  transposed  to  the 
last  member.  Of  course,  this  will  change  the  signs ;  and 
the  equation  will  become  ll^=a?.  And  if  ll-|=a;,  it  is 
evident  that  x=l\\.  This  coincides  with  what  was  shown 
in  §57. 

§59.  From  what  has  just  been  said,  we  see  that  all  the 
terms  of  each  member  may  be  transposed,  so  that  the  sign 
of  each  term  may  be  changed ;  and  still  the  equation  shall 
retain  the  same  members  as  at  first ;  and  that  it  is  also  im- 
material which  member  is  written  first.  And  hence,  in  any 
equation  the  signs  of  all  the  terms  may  he  changed  without 
affecting  the  equality. 

§60.  It  is  evident  that  all  the  terms  of  one  member  may 
be  transposed  to  the  other  member.  When  this  has  been 
done,  the  memberyrom  vyhich  the  terms  have  been  trans- 
posed becomes,  0.  Thus,  the  equation  x=^y — a^y,  may  be 
made  x-\-xy — 3i/=0  ;  where  — Sy  balances  x+xy. 

PROBLEMS. 

1.  A  man  has  six  sons,  whose  successive  ages  difl^er  by 
4  years ;  and  the  eldest  is  three  times  as  old  as  the  youngest. 
What  are  their  ages  1 


((           (( 

next 

u           n 

next 

((           (( 

next 

((           (( 

next 

((                   C( 

eldest. 

TRANSPOSITION  BY  ADDITION.  41 

Stating  the  question,  x=  age  of  the  youngest. 

ar+4= 
ar+4+4= 
ar+4  +  44-4= 
a?+4+4+4  +  4= 
ar+4+44-4+4+4= 
Forming  the  equation,  07+4  + 4+4  + 4  + 4=  3a; 
Uniting  terms,  ic+20=3a; 

Transposing  the  3x,  x — 3a? +24=0 

Transposing  20,  x — 3a7=  — 20 

Uniting  terms,  — 2x=  — 20 

Dividing  by  2,  — x= — 10 

or  a?=10 

2.  A  person  bought  two  horses,  and  also  a  hundred  dol- 
lar harness.  The  first  horse,  with  the  harness,  was  of 
equal  value  with  the  second  horse.  But  the  second  horse 
with  the  harness  cost  twice  as  much  as  the  first.  What 
was  the  price  of  each  horse  ? 

Stating  the  question,  x=  price  of  the  first. 

a;+100=  price  of  the  second. 
0?+ 100 +  100=  2d  horse  harnessed. 
Forming  the  equation,  a?+100  +  100=2a? 
Transposing  from  both  members,  x — 2a7= — 100 — 100 
Uniting  terms,  — x=  — 200 

Or  a?=200&c. 

3.  A  privateer  running  at  the  rate  of  10  miles  an  hour, 
discovers  a  ship  18  miles  ofl?*  sailing  at  the  rate  of  8  miles 
an  hour.  How  many  hours  can  the  ship  run  before  she 
will  be  overtaken  by  the  privateer  ? 

0^  The  equation  will  be  10a?=8a?+18.     Ans.  9  hours. 

4.  A  gentleman  distributing  money  among  some  poor 

4* 


48  ..      .  t  ALGEBRA.  [Eq.  SeC.  4. 

people,  found  that  he  would  lack  10  shillings  if  he  under- 
took to  give  5s  to  each.  Therefore,  he  gave  only  4s  to 
each,  and  finds  that  he  has  5s  left.  How  niany  persons 
were  there. 

0^  It  will  be  found  that  his  money  by  the  first  supposi- 
tion =bx — 10;  and  by  the  last  supposition,  it  =4a?+5. 

Ans.  15. 

5.  I  once  had  $84  in  my  possession ;  and  I  gave  away 
so  much  of  it,  that  I  have  now  three  times  as  much  as  I 
gave  away.     How  much  did  I  give  away  ? 

0^  If  I  gave  away  $a?,  then  $84--x  will  be  what  re- 
mains. Ans.  $21. 

6.  A  certain  sum  of  money  was  shared  among  five  per- 
sons, A,  B,  C,  D,  and  E.  Now,  B  received  $10  less  than 
A  ;  C  $16  more  than  B;  D  $5  less  than  C ;  E  $15  more 
than  D.  And  it  was  found  that  the  shares  of  the  last  two 
put  together,  were  equal  to  the  sum  of  the  shares  of  the 
other  three.     How  much  did  each  man  receive  1 

Ans.  A  $21 ;  B  $11  ;  C  $27  ;  D  $22  ;  E  $37. 

7.  A  person  wishes  to  give  3  cents  apiece  to  some  beg- 
gars, but  finds  that  he  has  not  money  enough  by  8  cents. 
He  gives  them  2  cents  apiece  and  has  Scents  left.  How 
many  beggars  were  there  1  Ans.  1 1 . 

8.  A  courier  who  had  started  from  a  certain  place  10 
hours  ago,  is  pursued  by  another  from  the  same  place,  and 
on  the  same  road.  The  first  goes  4  miles  an  hour,  and  the 
second  9.  In  how  many  hours  will  the  second  overtake  the 
first? 

Oy"  In  the  operation,  it  must  be  remembered  how  far  the 
first  had  the  start,  before  the  equal  time  for  both  began. 


MULTIPLICATION  OP  COMPOUND  QUANTITIES.  43 

MULTIPLICATION  OF  COMPOUND  QUANTITIES  BY 
SIMPLE  QUANTITIES. 

§61.  Suppose  you  purchase  8  melons  at  7  cents  apiece, 
and  afterwards  find  that  you  must  give  5  cents  apiece  more 
for  them.  In  this  case  you  pay  8  times  7  cents,  and  also 
8  times  5  cents  ;  that  is,  first,  56  cents,  and  afterwards  40 
cents. 

§62.  Let  us  apply  this  principle  to  Algebra.  You  pay 
in  all,  8  times  7  +  5,  which  =564-40.  Which  shows  that 
in  multiplying  a  compound  quantity,  you  multiply  each 
term  by  itself, 

■  We  can  easily  see  that  this  operation  will  give  the  right 
answer;  for  in  the  case  of  the  melons,  they  cost  12  cents 
apiece,  and  therefore  their  whole  cost  was  8  times  12  cents 
which  =96  cents.  But  the  answer  just  obtained,  56-f40 
=96. 

§63.  But  suppose  that  after  you  had  paid  7  cents  apiece, 
a  deduction  of  5  cents  apiece  was  made.  The  whole  cost 
would  then  be  8  times  7 — 5,  which  =56 — 40.  And  this 
agrees  with  the  truth  ;  for  you  first  paid  56  cents,  and 
afterwards  40  cents  were  deducted. 

§64.  This  shows  that  +  multiplied  by  -}-,  produces  +  j 
and  —  multiplied  by  +,  produces  — . 

EXAMPLES. 

1.  Multiply  a: 4-4,  by  3.  Ans.     Sx+U, 

2.  Multiply  124-ar,  by  5.  Ans.     604- 5a:. 

3.  Multiply  X— 10,  by  8.  Ans.     8a;— 80. 

4.  Multiply  126— a:,  by  4.  Ans.  504— 4ap. 

5.  Multiply  X+&,  by  6. 


H.. 


44  .^  ALGEBRA.  [Eq.  SeC.  5. 

t 

6.  Multiply  40  + a-,  by  10. 

7.  Multiply  a?— 32,  by  9. 

8.  Multiply  52— X,  by  12. 

9.  Multiply  2a;+ 14,  by  7. 

10.  Multiply  27  + 3a:,  by  14. 

11.  Multiply  3ar— 62,  by  15. 

12.  Multiply  97— 4a^,  by  12. 

13.  Multiply  a; +7 — y,  by  7. 

14.  Multiply  3a:+y— 12,  by  8. 

15.  Multiply  2x—Sy—e,  by  6. 

16.  Multiply  3x—l2+y,  by  5. 

§65.  Franciscus  Vieta,  a  Frenchman,  introduced  about 
the  year  1600,  the  vinculum  or  a  straight  line  drawn  over 
the  top  of  two  or  more  quantities  when  it  is  wished  to  con- 
nect them  together.  Thus,  a:  +  4x3,  signifies  that  both  x 
and  4  are  to  be  multiplied  by  3. 

§66.  In  1629,  Albert  Girard,  a  Dutchman,  introduced 
the  parenthesis  as  a  convenient  substitute,  in  many  cases, 
for  the  vinculum.  Thus,  (a: 4-4)  X  3,  isthe  sameasa:+4x3; 
and  is  read,  a?+4,  both  x3.  If  there  are  more  than  two 
terms  under  the  vinculum,  we  say,  afler  repeating  those 
terms,  a//,  &c.  Thus,  (ar-f-j^)  x(« — ^+c),  is  vead  x-\-y 
both  into  a — b-^-c  alL     See  also  §100. 

EQUATIONS. SECTION  5. 

1.  i^Xll=121,  to  find  a:. 
Solution,  a;— 9X11=121 

Performing  the  multiplication,  11a: — 99=121 
Transposing  and  uniting,  llar=220 

Dividing  by  11,  a:=20. 


MULTIPLICATION  OF  COMPOUND  QUANTITIES.  45 

2.  Given  (a?+7)x6=54,  to  find  x.  Ans.  x=2. 


3.  Given  12+a:x5=100,  to  find  x.  Ans.  8. 

4.  Given  x — 9x8=96,  to  find  ar.  Ans.  21, 


5.  Given  181— 3a:x5-=920,  to  find  x.       Ans.  61. 

6.  Given  (8+a;)x2-f  14=72,  to  find  x.     Ans.  21. 

7.  Given  (154-a;)x3— 27=48,  to  find  a?.    Ans.  10. 

8.  (112— 2a:)  x  3=  (2a:— 7)  X  4,  to  find  x.  Ans.  26. 

9.  (3a;+14)x4=(78— a-)x5,to  find  x,     Ans.  19-y\. 


10.  2jr-|-8x5=(324-x)x3,  tofinda:.  Ans.  8. 

11.  (3j:— 14)x7=(17— a:)x6,  to  find  x,      Ans.  14J. 


12.  120- 3arX2=  (4a:— 6)x9,  to  find  x,      Ans.  7. 

PROBLEMS. 

1.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money 
in  trade  ;  A  gains  $126,  and  B  loses  $87 ;  and  now  A's 
money  is  double  of  B's.     What  did  each  lay  out  1 

Staling  the  question,  ar,=  the  sum  for  each. 

a:+126=A's  sum  now. 
X — 87=  B's  sum  now. 
2a:— 174=  the  double  of  B's. 
Forming  the  equation,         a:  +  126=2a; — 174 
Transposing  and  uniting,         — x=  — 300 
Changing  signs,  ar=300  the  answer. 

2.  A  person,  at  the  time  he  was  married,  was  3  times  as 
old  as  his  wife;  but  after  they  had  lived  together  15  years, 
he  was  only  twice  as  old.  What  were  their  ages  on  their 
wedding  day  ? 


46  ALGEBRA.  [Eq.  SeC.  5. 

Stating  the  question,  x=  the  wife's  age. 

3x=  the  man's  age. 
ar-f-15=  the  wife's  after  15  years. 
3a?+15=  the  man's  after  15  years. 
2j:-f-30=  twice  the  wife's  age. 
Forming  the  equation,  3a: -j-15=2a?+<30 
Transposing  and  uniting,       a; =15  the  wife's  age. 
3a;=45  the  man's  age. 

3.  A  man  having  some  calves  and  some  sheep,  and  being 
asked  how  many  he  had  of  each  sort ;  answered  that  he 
had  twenty  more  sheep  than  calves,  and  that  three  times 
the  number  of  sheep  was  equal  to  seven  times  the  number 
of  calves.     How  many  were  there  of  each? 

0^  If  x==  number  of  calves,  there  iP+20=  number  of 
sheep.  Ans.  15  calves,  and  35  sheep. 

4.  Two  persons,  A  and  B,  having  received  equal  sums  of 
money,  A  paid  away  $25,  and  B  paid  away  $60  ;  and  then 
it  appeared  that  A  had  just  twice  as  much  money  as  B. 
What  was  the  sum  that  each  received?  Ans.  $95. 

5.  Divide  the  number  75  into  two  such  parts,  so  that 
three  times  the  greater  may  exceed  7  times  the  less  by  15. 

%CJ^  Ux=  the  greater,  thus  75 — x=  the  less  ;  and  Sx 
will  =7  times  the  less  4-15.  Ans.  54  and  21, 

6.  The  garrison  of  a  certain  town  consists  of  125  men, 
partly  cavalry  and  partly  infantry.  The  monthly  pay  of  a 
horse  soldier  is  $20,  and  that  of  a  foot  soldier  is  $15  ;  and 
the  whole  garrison  receives  $2050  a  month.  What  is  the 
number  of  cavalry,  and  what  of  infantry? 

10*^0;=  number  of  cavalry,  then  20a:=  their  whole 
pay,  &c.  Ans.  35  cavalry,  and  90  infantry. 

7.  A  grocer  sold  his  brandy  for  25  cents  a  gallon  more 
than  he  asked  for  his  wine ;  and  37  gallons  of  his  wine 


MULTIPLICATION  OF  COMPOUND  QUANTITIES.  Aj 

came  to  as  much  as  32  gallons  of  his  brandy.     What  was 
each  per  gallon  ? 

Ans.  $1.60  for  wine ;  and  $1.85  for  brandy. 

8.  A  wine  merchant  has  two  kinds  of  wine ;  the  one 
costs  9  shillings  a  gallon,  the  other  5.  He  wishes  to  mix 
both  wines  together,  so  that  he  may  have  50  gallons  that 
may  be  sold  without  profit  or  loss  for  8  shillings  a  gallon. 
How  much  must  he  take  of  each  sort  ? 

^3*  The  whole  mixture  will  be  worth  50  times  8  shillings. 
Ans.  37-^  gallons  of  the  best ;  and  12-|  of  the  poorer. 

9.  A  gentleman  is  now  40  years  old,  and  his  son  is  9  years 
old.  In  how  many  years,  if  they  both  live,  will  the  father 
be  only  twice  as  old  as  his  son  ? 

IC?"  In  X  years  he  will  be  40-f-a;,  and  his  son  9-\-x, 

Ans.  22  years* 

10.  A  man  bought  20  oranges  and  25  lemons  for$1.95» 
For  each  of  the  oranges  he  gave  3  cents  more  than  for  a 
lemon.     What  did  he  give  apiece  for  each  ? 

Ans.  3  cents  for  lemons,  6  cents  for  oranges. 

11.  A  man  sold  45  barrels  of  flour  for  8279  ;  some  at 
$5  a  barrel,  and  some  at  $8.  How  many  barrels  were  there 
of  each  sort?  Ans.  27  at  $5  ;  and  18  at  $8. 

12.  Says  John  to  William,  I  have  three  times  as  many 
marbles  as  you.  Yes,  says  William  ;  but  if  you  will  give 
me  20,  I  shall  have  7  times  as  many  as  you.  How  many 
has  each  ? 

ICT*  Let  x=  William's  and  3x=  John's.  Then  after  the 
change,  a;+20=  William's  and  3a; — 20=  John's. 

Ans.  John  24  ;  William  8. 

13.  A  person  bought  a  chaise,  horse,  and  harness,  for 
$440.     The  horse  cost  him  the  price  of  the  harness,  and 


48  ALGEBRA.  [Eq.  SeC.  5. 

$30  more  ;  and  the  chaise  cost  twice  the  price  of  the  horse. 
What  did  he  give  for  each  ? 

Ans.  For  the  harness  $50  ;  horse  8130;  chaise  $260. 

14.  Two  men  talking  of  their  ages,  the  first  says  your 
age  is  18  years  more  than  mine,  and  twice  your  age  is 
equal  to  three  times  mine.     What  is  the  age  of  each  1 

Ans.  Youngest  36  years.     Eldest  54  years. 

15.  A  boy  had  41  apples  which  he  wished  to  divide 
among  three  companions  as  follows ;  to  the  second,  twice  as 
many  as  to  the  first,  and  3  apples  more ;  and  to  the  third, 
three  times  as  many  as  to  the  second,  and  2  apples  more. 
How  many  did  he  give  to  each  ? 

Ans.  To  the  first  3  ;  second  9  ;  third  29. 

16.  How  many  gallons  of  wine,  at  9  shillings  a  gallon, 
must  be  mixed  with  20  gallons  at  13  shillings,  so  that  the 
mixture  may  be  worth  10  shillings  a  gallon  ? 

Ans.  60  gallons. 

17.  Two  persons,  A  and  B,  have  each  an  annual  income 
of  $400.  A  spends,  every  year,  $40  more  than  B  ;  and, 
at  the  end  of  4  years,  they  both  together  save  a  sum  equal 
to  the  income  of  either.     What  do  they  spend  annually? 

Ans.  A  $370  ,•  B  $330. 


ii-  arff  h 


FRACTIONS.  49 


FRACTIONS. 


§65.  All  the  division  which  the  pupil  has  as  yet  perform- 
ed, has  been  the  division  either  of  numeral  quantities,  or  of 
the  numeral  co-efficients.  But  in  Algebra,  it  is  frequently- 
necessary  to  divide  literal  quantities.  For  example,  after 
having  made  x  to  stand  for  an  unknown  quantity,  we  may 
wish  to  find  the  half  of  x,  or  the  third  of  re,  or  the  fourth 
of  X,  &c. 

§66.  In  common  arithmetic,  if  we  wish  to  divide  1  by  2, 
we  do  it  by  writing  2  under  the  1 ;  thus,  ^.  So  if  we  wish 
to  divide  2  by  3,  we  write  3  under  the  2  ;  thus,  |.  In  the 
same  manner,  2  divided  by  5  is  written  f  ;  3-^-4  is  writ- 
ten I ;  6-i-7  is  written  f.  The  quantities  that  are  obtain- 
,ed  by  dividing  in  this  manner,  are  caWed  fractions, 

§67.  In  Algebra,  we  most  generally  make  use  of  this 
method  of  dividing ;  especially  when  we  divide  literal 
.quantities.  Or,  in  other  words,  we  divide  a  literal  quan- 
tity by  writing  the  divisor  under  the  dividend,  with  a 
straight  line  between  them  ;  thus,  x  divided  by  2,  is  written 
~;  and  is  read,  x-half.  x-r-S,  is  written  |;  and  is  read, 
c-third  ;  a:-i-4,  is  written  J,  and  is  read  xfourth  ;  3a:-T-4, 
J,  and  is  read,  Sx fourth. 

§68.  The  two  separate  numbers  that  we  employ  in  writ- 
ng  a  fraction,  are  called  terms.  The  upper  term  is  called 
he  numerator,  and  the  lower  term  is  called  the  denomina- 
or.  Thus,  in  the  fraction  -j,  we  call  x  the  numerator,  and 
J  thC' denominator. 

§69.  If  the  one-third 'of  x  is  |,  it  is  evident  that  |  of  x, 
i  two  times  as  much ;  that  is  y.     If  -}•  of  x  is  ^,  then  ^  o^ 
5 


50  ALGEBRA.  [Eq.  SeC  5. 


X  is  — •     Whence  the  rule  to  multiply  a  whole  number  by  a 


5 

fraction,  is,  to  multiply  the  whole  number  by  the  numerator y 
and  divide  by  the  denominator  ;  as  J  of  a:  is  y ;  f-  of  i/  is 

%.  2  of  a  is-. 
—,   3-01  a  ih  3  . 

Examples.  The  pupil  may  multiply  a,  x,  and  y,  each 
of  them  by  | ;  and  then  by  J ;  and  then  by  f ,  |,  |,  |,  f,  ^, 
successively. 

^70.  As  we  can  multiply  a  number  of^ parts  as  well  as  a 
number  of  wholes,  and  as  the  denominator  is  nothing  more 
than  the  name  of  the  parts  ;  it  is  plain,  that  to  multiply  a 
fraction,  we  multiply  the  numerator,  and  retain  the  deno- 
minator without  alteration.  Thus,  2  times  y  is  |^ ;  3  times 
5  is  1/  ;  2  times  ^  is  y  ;  4  times  =|  is  ^|'  &c. 

Examples.  Multiply  each  of  the  following  fractions  by 
2,  then  by  3,  and   then   by  4.     |,  f ,  f  4^  e,  |,  |,  |,  |, 

21    3i    2z    3x.  41 
3'T'    5"    5      7 

§71.  As  we  know  that  2  halves  =  a  whole,  we  readily 
conclude  that  4  halves  =  2  wholes ;  and  that  6  halves  =  3 
wholes;  &c.  Likewise,  because  3  thirds  =  a  whole,  6 
thirds  must  equal  2  wholes;  and  9  thirds  must  equal  3 
wholes.  In  the  same  manner  8  fourths  =  2  wholes ;  20 
fifths  «  4  wholes ;  18  thirds  =  6  wholes  ;  &c.  Such  frac- 
tions are  called  improper  fractions. 

§72.  Hence,  in  order  to  find  how  many  whole  ones  there 
are  in  any  number  of  halves,  we  have  only  to  see  how 
many  times  two  halves  are  contained  in  that  number. 
Thus,  in  10  halves  there  are  as  many  whole  ones  as  there 
are  2  halves  contained  in  10  halves ;  which  is  5.  In  the 
same  manner,  in  12  thirds  there  are  as  many  whole  ones 
as  there  are  3  thirds  contained  in  12  thirds;  which  is  4. 

§73.  Thus  we   have  the  rule,  to  change  an  improper 


FRACTIONS.  51 

fraction  to  a  whole  number,  divide  the  numerator  by  the 
denominator. 

When  the  answer  consists  of  an  integer  and  a  fraction, 
it  is  called  a  mixed  number. 

Examples,  1.  How  many  whole  ones  in  |. 

Ans.  8-T-3=:*4. 

2.  How  many  whole  ones  in  J  ?   ^^  ?  Y  ^  \^  ^  ¥  • 

3.  How  many  whole  ones  in  ^i  ?  ^  ?  ^^  1  -3/  ?  \o  ? 

4.  How  many  whole  a:'s  in  ^1  ^?  ?^"?  ^? 

.  How  many  whole  xs  i"  -^  •  '^3'  •   T     "7 

6.  How  many  whole  x's  in  3  times  ^? 

7.  How  many  whole  x's  in  4  times  1 1 

8.  How  many  whole  x's  in  5  times  —  ? 

5 

§74.  If  we  have  the  quantity  f ,  we  know  that,  as  it 
takes  5  fifths  to  make  a  whole  one,  it  will  take  5  times  this 
quantity  to  make  a  whole  x.  Therefore,  if  we  multiply  ^ 
by  5,  we  shall  obtain  — ,  or  exactly  ;r.  If  we  multiply  * 
by  3,  we  shall  obtain  — ,  or,  which  is  the  same,  x.  If  we 
multiply  5  by  4,  we  shall  obtain  —,  or  x. 

§75.  As  4  times  ^  is  equal  to  x\  then  4  times  -^  must 
be  equal  to  twice  as  much,  or  2x ;  and  4  times  -^  must  be 
three  times  as  much,  or  ^x.  As  3  times  f  is  equal  to  x  ; 
so  3  times  ^  must  be  twice  as  much,  or  2x.  So  5  times -^ 
must  be  3  times  as  much  as  5  times  y ;  and  therefore  is  3ar. 

§76.  'Any  fraction  when  multiplied  by  the  number  which 
is  the  same  as  the  denominator,  will  produce  a  quantity 


52  ALGEBRA.  [Eq.  SeC.   5 

which  is  the  same  as  the  numerator.     Thus, 

We  shall  be  able  to  make  use  of  this  principle  in  the 
solution  of  many  equations,  if  we  operate  in  accordance 
with  the  following  axiom  or  self-evident  truth. 

§77.  If  equals  be  multiplied  by  the  same,  their  products 
will  be  equal  Thus,  if  a:=10,  then  2a:=20  ;  4a:=:40  ; 
&c. 


EaUIVALENT  FRACTIONS. 

§78.  It  is  evident  [§71,]  that  each  of  the  following  fractions, 
f '  h  J»  T'  h  h  h  h  *^^'»  '^^  equal  to  1.  Therefore,  they 
must  be  equal  to  one  another.  Also,  each  of  the  following 
fractions,  f,  |,  4,  |,  y,  &c.,  is  equal  to  2  ;  and  conse- 
quently they  are  all  equal  to  one  another.  In  the  same 
manner,  we  may  make  many  fractions  that  will  equal  3  ; 
and  so  of  any  other  number. 

§79.  Let  us  take  from  the  first  set  of  the  above  fractions, 
I  and  I  which  are  equal  to  one  another.  We  see  that  both 
the  numerator  and  denominator  in  the  last  fraction  are  twice 
as  much  as  in  the  first.  We  see  the  same  fact  in  the  equal 
fractions  -J  and  J  ;  and  also  in  the  equal  fractions  ^  and  |. 
We  find  the  same,  by  taking  from  the  second  set,  the  equal 
fractions  f  and  |  ,•  and  also  |  and  | ;  and  also  f  and  y . 

§80.  Again  in  the  equal  fractions,  -|  and  |,  we  find  each 
term  in  the  last  fraction  three  times  as  great  as  the  corre- 
spondent term  in  the  first  fraction.  The  same  may  be  ob- 
served in  the  fractions  |  and  | ;  and  also  in  f  and  -f  ,*  and 
also  in  |  and  |. 


EQUIVALENT  FRACTIONS.  53 

§81.  By  pursuing  this  investigation,  we  shall  find  that 
whenever  we  multiply  both  the  numerator  and  the  denomi- 
nator  by  the  same  number,  no  matter  what  tliat  number  may 
be,  the  fraction  made  by  that  multiplication,  will  be  equal 
to  the  first  fraction.  Hence,  there  is  an  equality  between 
the  following  fractions,  |,  f ,  f,  |,  -^-q^  -j-%,  &;c.;  and  also  be- 
tween the  following,  ^,  |,  |,  y\.  /y,  y\,  &c. 

§82.  The  principle  just  explained,  leads  to  another  which 
is  of  much  importance.  Suppose  we  multiply  by  8,  both 
terms  of  the  fraction  | ;  and  obtain  -|-|.  It  is  plain  that 
both  terms  of  the  fraction  }f  can  be  divided  by  8,  to  bring 
the  fraction  back  tof.  So  also,  if  both  terms  in  ^  be  mul- 
tiplied by  6,  the  fraction  will  be  H,  which  means  just  as 
much  as  | ;  and,  of  course,  if  both  terms  in  J|  be  divided 
by  6,  the  fraction  will  be  brought  back  to  |,  which  is  equal 
to  ^|.  So,  in  general,  if  we  divide  both  the  numerator 
and  the  denominator  of  a  fraction  by  the  same  nuTnber,  we 
have  a  new  fraction  which  will  be  equal  to  the  first.  Thus, 
j%  may  be  changed  to  | ;       jf  to  | ;       ^{  to  f . 

§83.  It  is  evident,  that  of  several  fractions  of  equal  value, 
that  which  has  the  least  denominator  is  the  most  easily  un- 
derstood. Thus,  I  of  an  apple  is  much  better  known  at 
first  sight,  than  ||  of  an  apple.  And  ivhen  a  fraction  is 
brought  to  as  small  a  denominator  as  it  can  be  changed  to j 
we  say  it  is  reduced  to  its  lowest  tcr?ns. 

§84.  In  order  to  reduce  a  fraction  to  its  lowest  terms,  di- 
vide both  the  numerator  and  the  denominator  by  any  number 
that  will  divide  each  without  a  remainder*  Thus,  in  the 
fraction  -j^t'  ^^^^  terms  may  be  divided  by  5,  by  which  we 
obtain  |A ;  and  both  terms  of  this  last  fraction  may  be  di- 
vided by  8,  by  which  we  obtain  |-. 


54  ALGEBRA.  [Eft.  SeC.  6* 

EaUATIONS. — SECTION  6. 

Problems. 
1.  In  an  orchard,  1  of  the  trees  bear  apples,  |  of  them 
bear  pears,  ^^  bear  plums,  and  81  bear  cherries.     How 
many  trees  are  there  in  the  orchard  ;  and  how  many  of  each 
sort  1 

Stating  the  question,  a;=  number  of  trees. 

1=  apple  trees. 

y  =  pear  trees. 

-    ,  ■^y=  plum  trees. 

81=  cherry  trees. 
All  these  trees  together  =  the  whole  orchard* 
Forming  the  equation,  |4-|  +  ^£-f  81s=a;. 

Now,  we  know  that  if  we  multiply  ^  by  4  we  obtain  x 
alone  ;  that  is,  we  destroy  the  fraction,  and  make  it  a  whole 
number.  And  we  know,  that  if  we  multiply  the  first  mem- 
ber by  4 ;  and  also  the  last  member  by  4,  we  shall  not  de- 
stroy the  equation.  See  §77.  We  will  therefore  multiply 
both  members  by  4,  for  the  purpose  of  destroying  the/rac- 
tion  in  the  first  term.  It  will  then  become 
a:+T+^+324=4a;. 

Next,  we  will  multiply  both  members  by  5,  to  destroy  the 
fraction  in  the  second  term.     This  will  make 
5a;4-4a:+^''+1620=20a:. 

Then  we  will  multiply  by  11,  to  destroy  the  remaining 
fraction,  which  will  make 

55a;+44a;+40a;+17820=220a?. 
Transposing  and  uniting,  *-81a:= — 17820. 

Changing  signs,  81a?=sl7820. 

Dividing  by  81,  a;  =220.    the  Ans. 


'  «c' 


EaUIVALENT  FRACTIONS.  55 

2.  In  a  certain  school,  |  of  the  boys  learn  mathematics, 
A  of  them  study  Latin  and  Greek,  and  6  study  English 
grammar.     What  is  the  whole  number  of  scholars? 

0:^  After  the  question  is  stated,  the  equation  will  be- 
come ?+!''+ 6=  a: 
Multiplying  by  5,  a:+^+30=5a: 
Multiplying  by  4,  4a; + 15a; +120=  20a: 
Transposing  and  uniting,  — a:= — 120.    Ans. 

3.  A  gentleman  has  an  estate,  -}  of  which  is  woodland? 
f  of  it  pasture,  and  105  acres  embrace  the  pleasure  grounds, 
gardens,  and  orchards.     How  many  acres  does  it  contain  1 

Ans.  630  acres. 

4.  After  paying  away  i  and  ^  of  my  money,  I  find  22 
dollars  yet  in  my  purse.     How  much  had  I  at  first  ? 

Ans.  $A0. 

5.  A  man  bought  a  lot  of  ground,  for  which  he  agreed 
to  pay  as  follows  :  i  of  the  money  on  taking  possession,  -J 
of  it  in  6  months,  and  $250  at  the  end  of  the  year.  How 
much  did  he  pay  in  all  ?  Ans.  $600. 

6.  A  post  is  one-fourth  of  its  length  in  the  mud,  one-third 
in  the  water,  and  10  feet  abov^e  the  water.  What  is  its 
whole  length  ?  Ans.  24  feet* 

1.  In  a  Christmas  pudding,  ^  is  flour,  \  milk,  -J-  eggs,  \ 
suet  and  fruit,  and  J  of  a  pound  of  spices  and  other  ingre- 
dients.    What  is  the  weight  of  the  pudding  1 

{^  The  equation  will  be  l+f +J4-?+J-^-ar. 

Ans.   15  pounds* 

8.  A  lady  being  asked  what  her  age  was  ;  replied,  if  you 
add  ^,  i,  and  ^  of  my  age  together,  the  sum  will  be  18« 
How  old  \yas  she  ? 


I- 


56        ^  ALGEBRA.  [Eq.  SeC.  6. 

Cfr  After  the  question  has  been  stated,  the  equation  will 
be,     |-f  £_j-|=l8.  Ans.  24  years. 

9.  What  sum  of  money  is  that,  whose  ^  part,  i  part, 
and  -J-  part,  added  together,  amount  to  94  dollars  1 

Ans.  8120. 

10.  A  person  found  upon  beginning  the  study  of  his  pro- 
fession, that  he  had  passed  -f  of  his  life  before  he  com- 
menced his  education,  ^  of  it  under  a  private  teacher,  the 
same  tirne  at  a  public  school,  and  four  years  at  the  univer- 
sity.    What  was  his  age?  Ans.  21  years. 

11.  How  much  money  have  I  in  my  pocket,  when  the 
fourth  and  fifth  part  of  it  together,  amount  to  $2.25  ? 

Ans.  $5. 

12.  The  3d  part  of  my  income,  said  a  person,  I  expend 
in  board  and  lodging,  the  8th  part  of  it  in  clothes  and  wash- 
ing, the  10th  part  of  it  in  incidental  expenses,  and  yet  I 
save  8318  a  year.     What  was  his  yearly  income  ? 

Ans.  8720. 

13.  A  gentleman  bequeaths,  in  his  will,  the  half  of  his 
property  to  his  wife,  one-sixth  part  to  each  of  his  two  sons, 
the  twelfth  part  to  his  sister,  and  the  remaining  8600  to  his 
servant.     What  was  the  amount  of  his  property? 

Ans.  87200. 

14.  Of  a  piece  of  metal,  -J  plus  24  ounces  is  brass,  and 
.|  minus  42  ounces  is  copper.  What  is  the  weight  of  the 
piece? 

0^  The  equation,  when  formed,  will  be 

l+24-f^— 42=ar. 

Ans.  216  oz. 

15.  A  farmer  mixes  a  quantity  of  grain,  so  that  20 
bushels  less  than  |-  of  it  is  barley,  and  36  bushels  more 


EQUIVALENT  FRACTIONS.  57 

than  ^  of  it  is  oats.     How  many  bushels  are  there  in  the 

whole;  and  how  many  of  each  sort? 

0^  In  stating  the  question  ;  ^ — 20=  barley,  and  ^■+ 

36=  oats. 

Ans.  96  bushels  in  all ;  28  of  barley,  and  68  of  oats. 

16.  A  teacher  being  asked  how  many  scholars  he  had, 

replied.  If  I  had  as  many  more,  half  as  many  more,  and 

quarter  as  many  more,  I  should  have  88.     How  many  had 

he? 
|C?^  In  stating  the  question,  he  has  x ;  and  as  many 

more  is  another  x  ;  &c.  Ans.  32. 

17.  In  a  mixture  of  copper,  tin,  and  lead ;  16  pounds 
less  than  2  was  copper,  12  pounds  less  than  ^-  was  tin,  and 
4  pounds  more  than  i  was  lead.  What  was  the  weight  of 
the  whole  mixture ;  and  also  of  each  kind  ? 

Ans.  2881b.;  and  also  1281b.,  841b.,  and  761b. 

18.  What  is  that  number  whose  -J  part  exceeds  its  -J  part, 
by  12? 

iHj^  The  statement  is  the  same  as,  ^  of  it  equals  i  of  it 
4-12.  Ans.  144. 

19.  What  number  is  that  whose  A  part  exceeds  its  -}  part 
by  72  ?  Ans.  540. 

20.  A  certain  sum  of  money  is  to  be  divided  amongst 
three  persons.  A,  B,  and  C,  as  follows.  A  is  to  receive 
83000  less  than  half  of  it,  B  81000.  less  than  the  third  of  it, 
and  C  8800  more  than  the  fourth  of  it.  What  is  the  sum 
to  be  divided  ;  and  what  does  each  receive? 

Ans.  838400;  and  also,  816200,  8II8OO,  810400. 

21.  A  man  driving  his  geese  to  market,  was  met  by  an- 
other, who  said.  Good  morrow,  master,  with  your  hundred 
geese.     He  replied,  I  have  not  a  hundred  ;  but  if  I  had  as 


-'^S  ALGEBRA.  [Eq.  SeC.  6. 

many  more,  and  half  as  many  more,  and  two  geese  and  a 
half,  I  should  have  a  hundred.     How  many  had  he? 

Ans.  39  geese. 

22.  A  shepherd,  being  asked  how  many  sheep  he  had, 
replied.  If  I  had  as  many  more,  half  as  many  more,  and  7 
sheep  and  a  half,  I  should  have  just  500.  How  many  sheep 
had  he?  Ans.  197  sheep. 

23.  A  legacy  of  $1200  was  left  between  A  and  B,  in 
such  a  manner,  that  ^  of  A's  share  was  equal  to  |  of  B's. 
What  sum  did  each  receive  ?  Ans.  A  $640  ;  B  $560. 

24.  If  the  half,  third,  and  fourth  parts  of  my  number, 
be  added  together,  the  sum  will  be  one  more  than  my  num- 
ber.    Now,  what  is  my  number?  Ans.  12. 

25.  A  says  to  B,  your  age  is  twice  and  ■§■  of  my  age ; 
and  the  sum  of  our  ages  is  54  years.  What  is  the  age  of 
each?  Ans.  A's  15  years;  B's  39  years. 

26.  A  young  gentleman  having  received  a  fortune,  spent 
■J  of  it  the  first  year,  ^  of  it  the  second,  and  -J-  of  it  the 
third,  when  he  had  $2600  left.  What  was  his  whole  fpr- 
tune? 

OO"  In  stating  the  question,  what  was  spent,  =  the  whole 
minus  $2600.  Ans.  $12000. 

27.  A  father  leaves  four  sons  who  share  his  property  in 
the  following  manner.  The  first  takes  half,  minus  $3000  ; 
the  second  lakes  a  third,  minus  $1000  ;  the  third  takes  ex- 
actly a  fourth;  and  the  fourth  son  takes  a  fifth  and  $600. 
What  was  the  whole  fortune,  and  what  did  each  son  receive  ? 

Ans.  The  whole  fortune  was  $12000,  and  each  son 
received  $3000. 


DIVISION  OF  COMPOUND  QUANTITIES.  59 


DIVISION  OF  COMPOUND  QUANTITIES  BY  SIMPLE 
QUANTITIES. 

§86.  We  have  found,  §67,  that  the  algebraical  method 
of  dividing,  is  to  write  the  divisor  under  the  dividend,  with 
a  straight  line  between  them.  It  is  plain  that  compound 
quantities  may  be  divided  in  this  way,  as  well  'as  simple 
quantities.  Thus,  14+x  is  divided  by  3,  as  follows: 
14+ar. 
3~ 

§87.  With  the  same  reason,  we  find  the  fraction  of  a 

compound  number,  by  multiplying  it  by  the  numerator,  and 

writing  the  denominator  under  the  product.     Thus, 

2    ^        ^  .    2x — 10. 

4  of  X — 5  IS  

^  3 

EXAMPLES. 

1.  What  is  J  of  84-a;  ?  Ans.  24  +  3ar. 


2.  What  is  I  of  ar— 27  ? 


3    What  is  I- of  3a:— 141 

4.  What  is  I- of  9  +  5.T? 

5.  What  is  ^  of  7ic— 19? 

6.  What  is  |  of  9a!— 27  1 

§88.  This  may  be  changed  into  whole  numbers  by  §73. 
Thus,  15f=l!£=5x-15. 


4 

Ans. 

2ar— 54. 

5 

Ans, 

.  6a:— 28. 

7 

Ans. 

27  + 15a-. 

8 

Ans. 

28a:-^76. 

7 

Ans. 

45a:— 135. 

60                                                     ALGEBRA.  [Eq.  SeC.  7. 

EQUATIONS. SECTION  7. 

Problems. 

1.  A  young  gentleman  being  asked  his  age,  said  it  is 
such  that  if  you  add  8  years  to  it,  and  then  divide  by  3, 
the  quotient  would  be  9.     How  old  was  he  ? 

Stating  the  question,  x=  his  at^e. 

a:+8=  with  8  added.  >;*': 
Forming  the  equation,  ^  *"   —9 

o 

Muhiplying  by  3,  a?+8=27 

Transposing  and  uniting,  a?=19  the  Ans. 

2.  A  man  being  asked  what  he  gave  for  his  horse,  re- 
plied, that  if  he  had  given  $12.  more,  |  of  the  sum  would 
be  84.     What  was  the  price  of  the  horse  ? 

Stating  the  question,  a?=  the  price. 

ic-t-12=  when  increased. 

- — ! —  s=|  of  the  sum. 

Forming  the  equation  ^' =84 

4 

Multiplying  by  4,  3a; -f  36=336 

Transposing  and  uniting,  3x=300 

Dividing  by  3,  a;=100  the  Ans. 

'3.  What  sum  of  money  is  that,  from  which  85  being 
subtracted,  two-thirds  of  the  remainder  shall  be  $40? 

Ans.  $65. 

4.  It  is  required  to  divide  a  line  that  is  15  inches  long 

into  two  such  parts,  that  one  of  them  may  be  |  of  the  other. 

0^  In  slating  the  question,  the  parts  are  x,  and  lo — x. 

Ans.  84;  and  6-f. 


DIVISION  OF  COMPOUND  QUANTITIES.  61 

5.  It  is  required  to  find  a  number,  such  that  if  15  be 
subtracted  from  it,  -J  of  the  remainder  shall  be  100  ? 

Ans.  140. 

6.  Divide  the  number  46  into  two  parts,  so  that  when 
one  is  divided  by  7,  and  the  other  by  3,  the  quotients  toge- 
ther may  amount  to  10.  Ans.  28  and  18. 

7.  A  person  being  asked  the  time  of  day,  answered  that 
the  time  past  from  noon  was  equal  to  |  of  the  time  to  mid- 
night.    What  was  the  hour  ? 

}Cj^  In  stating  the  question,  the  parts  are  x,  and  12 — x, 

Ans.  20  minutes  after  5. 

8.  Two  men  talking  of  their  horses,  A  says  to  B,  my 
horse  is  worth  $25  more  than  yours  ;  and  |-  of  the  value  of 
my  horse  is  equal  to  J  of  the  value  of  yours.  What  is  the 
value  of  each  ? 

iCj^  The  values  will  be  x^  and  a?+25. 

Ans.  A's  $125  ;  B's  100. 

9.  Two  persons  have  equal  sums  of  money.  One  hav- 
ing spent  $39,  and  the  other  $93,  the  last  has  but  half  as 
much  as  the  first.     How  much  had  each  ?  Ans.  $147. 

10.  A  man  being  asked  the  value  of  his  horse  and  chaise, 
answered  that  the  chaise  was  worth  $50  more  than  the 
horse;  and  that  one  half  the  value  of  the  horse  was  equal 
to  one  third  the  value  of  the  chaise.  AVhat  was  the  value 
of  each  ?  Ans.  Horse  $100 ;  Chaise  $150. 

11.  What  number  is  that,  to  which  if  I  add  13,  and  from 
Y^-j  of  the  sum  subtract  13,  the  remainder  shall  be  13? 

Ans.  825. 

12.  A  man  being  asked  the  value  of  his  horse  and  saddle, 
answered  that  his  horse  was  worth  $114  more  than  his  sad- 

6 


62  ALGEBRA.  [Eq.  SeC.  7. 

die,  and  that  |  of  the  value  of  his  horse  was  7  times  the 
value  of  his  saddle.     What  was  the  value  of  each  ? 

Ans.  Saddle  $12;  Horse  $126. 

13.  A  person  rented  a  house  on  a  lease  of  21  years,  and 
agreed  to  do  the  repairs  when  |  of  that  part  of  the  lease 
which  had  elapsed,  should  equal  ^  of  the  part  to  come. 
How  long  will  he  hold  possession  before  he  repairs  ? 

Ans.  12  years. 

14.  What  number  is  that,  to  which  if  I  add  20,  and 
from  f  of  this  sum  subtract  12,  the  remainder  shall 
be  10?  Ans.  13. 

15.  A  person  has  a  lease  for  99  years ;  and  being  asked 
how  much  of  it  was  already  expired,  he  answered  that  |  of 
the  time  past  was  equal  to  |  of  the  time  to  come.  What 
time  had  already  past  ?  Ans.  54  years. 

16.  Divide  $183  between  two  men,  so  that  ^  of  what  the 
first  receives,  shall  be  equal  to  -^^  of  what  the  second  re- 
ceives.    What  will  be  the  share  of  each  ? 

Ans.  $6.3  and  $120. 

17.  Bought  sheep  for. $300,  calves  for  $100,  and  pigs 
for  $25,  and  then  laid  out  $15,  which  was  f  of  the  rest  of 
my  money,  in  getting  them  home.  How  much  had  I  at 
first?  Ans.  $443. 

18.  A  gentleman  paid  four  labourers  $136.  To  the  first 
he  paid  three  times  as  much  as  to  the  second  wanting  $4  ; 
to  the  third  one  half  as  much  as  to  the  first,  and  $6  more; 
and  to  the  fourth  four  times  as  much  as  to  the  third,  and  $6. 
more.     How  much  did  he  pay  to  each  ? 

Ans.  To  the  first  $26 ;  second  $10 ;  third  $19;  fourth  $81. 


DIVISION  OF  FRACTIONS.  63 


DIVISION  OF  FRACTIONS,  AND  FRACTIONS  OF 
FRACTIONS. 

§90.  It  was  shown  in  §70,  that  in  multiplying  a  fraction, 
we  multiply  the  numerator  only,  and  retain  the  denomina- 
tor. On  the  same  principle,  a  fraction  is  divided  by  di- 
viding its  numerator,  and  retaining  its  denominator. 

Thus, --^3=-.    _^4=-,&c. 

§91.  But,  supposing  we  wish  to  divide  f  by  3.  In  this 
case,  we  cannot  divide  the  numerator  2  by  3  without  a  re- 
mainder ;  and  therefore  we  must  look  for  some  other  prin- 
ciple to  assist  us.  We  shall  find  it  in  §81,  where  it  was 
shown  that  a  fraction  may  be  changed  to  one  with  different 
terms,  without  altering  the  value. 

§92.  It  is  evident  then,  that  we  have  only  to  change  the 
fraction  which  is  to  be  divided,  to  some  equivalent  fraction, 
whose  numerator  can  be  divided  by  the  divisor  without  a 
remainder.  Thus,  f  can  be  changed  ^^■^,  if,  ||,  &c. ;  each 
of  which  can  be  divided  by  3,  giving  for  the  quotient  either 
yV  or  ^,  or  /j,  &c. 

§93.  The  most  convenient  equivalent  fraction  will  be  ob- 
tained by  multiplying  both  terms  of  the  fraction  by  the 
number  which  is  to  be  the  divisor.  Because  it  is  certain 
that  after  the  numerator  has  been  multiplied  by  a  number, 
the  product  can  in  return  be  divided  by  that  number. 


TK       4      ^      20  .20     ^       4 

Thus,  --^5==---r-5;  and  —-^5,=—. 
9  45  45  45 

§94.  But,  by  examining  the  example  just  given,  we  find 
the  numerator  of  the  answer  to  be  the  same  as  the  numera- 
tor of  the  first  fraction  ;  for  the  first  numerator  has  been 
multiplied   and  then  divided  again  by  the  same  number. 


64  ALGEBRA.  [Eq.  SeC.  7. 

The  denominator  only  is  changed ;  and  that  has  been  done 
by  multiplying  the  first  denominator  by  the  number  that 
was  to  be  the  divisor. 

§95.  Hence  the  rule  for  dividing  a  fraction.  Divide  its 
numerator  when  it  can  be  done  without  a  remainder.  But 
if  there  should  be  a  remainder,  multiply  the  denominator 
by  the  divisor  for  a  new  denominator  ;  and  leave  the  nu- 
merator as  it  is. 


Thus, 

EXAMPLES. 

^-4- 

1.  Divide 

>5. 

3a; 

2.  Divide 

->y- 

2a; 
Ans.-. 

3.  Divide 

-,^y^- 

Ans.  -. 

4.  Divide 

fb.5. 

Ans.f. 

5.  Divide 

-,^.,3. 

4a: 
Ans.    — . 
o 

6.  Divide 

'r^y- 

Abs.      g    . 

i.  Divide 

^s,. 

A                ^—^ 

^"'-     20.  • 

8.  What  is  1  of  ^t 

A        3a- 
^"«    28- 

At 
9.  What  is  i  of  y? 

A        4^ 
Ans.  -g. 

10.  What  is  i  of  llzf£? 

41— 2x 
Ans.       ,2 

DIVISION  OF  FRACTIONS.  65 


11. 

What  is  1  of'"   ;;-^2/, 

Ans. 

Sx—^l+y 
64 

12. 

What  is  i  of  '^-^1 

^           X — 1 

Ans. 

7a:  4-4 
5a:— 5' 

13. 

Whatis|of5gf? 

Ans. 

21— 3a: 
14a:— 63 

14. 

Whatisfoff+!!^ 

'  Ans. 

484-7y 

1  S'y 1  9/11 

|C7*  In  this  example,  although  we  can  divide  the  first 
term  in  the  numerator,  we  do  not  do  it  because  we  cannot 
divide  the  last  term. 

15.  What  is  i  of  -21=^1  Ans.  ,r^^=^ 

,«    TTn      •    ,     n  32a:  4-16^  ,  4a:  +  2 

16.  What  IS  I  of  ^^^  ?  Ans.    —^—. 

^        X — 6+2/  X — 64-1/ 

§96.  We  are  now  enabled  to  find  a  fraction  of  a  fraction 
by  the  rule  in  §69 ;  which  is  to  multiply  by  the  numerator, 
and  divide  by  the  denominator.  To  multiply  by  the  nume- 
rator is  to  multiply  the  numerators  together.  And  to  di- 
vide by  the  denominator,  we  have  just  shown,  i^  to  multiply 
the  denominators  together. 

EXAMPLES. 

1.  What  is  f  of  -  ?  Ans.—-  =— . 

^6  18       9 

2.  What  is  f  of  ^"]  Ans.^. 

3.  What  is  I  of  -  ?  ^"^'W* 

4.  What  IS  f  of  — -—  ?  Ans. 


21 


6* 


•^ 


#y 


66 


ALGEBRA. 


5.  What  is  I  of  Il±?f? 

6x 

6.  What  is  4  of  — —  ? 

^        23+2/ 


7.  What  is  j\  of 


91/ 


a;+61 


Ans. 
Ans. 
Ans 


[Ea.  Sec.  8. 
2734-6iC 


20 
24ic 


2074-91/ 
ISy 


Q.WhatiSf^of^f-^? 
^^       41 +6iC 

9.  What  is  ^  of  |^±i^? 
^       2ic— 112/ 


Ans. 


;^^^^      —~"llic+671 
36ar— 180     18a;~90 


10.  What  is  3?^  of 


6x+ly— 10, 
4  +  3^— 2ar 


11..  What  is  I  of  ^     ^     ^r 


12.  What  is  I-  of 


6  +  x—y 
Sx-f-5y—8^ 
27— 4a; 


410  +  60a;     205  +  30ar 
l2y-]-2Sx 
14a? — 772/* 
18a:+2l2/— 30 
40+302/— 2.0a;* 

4a; — 42/ — 24 
54+9a;— 9y* 

6a;+102/— 16 
135— 20a; 


Ans. 
Ans. 
Ans. 
Ans. 


EQUATIONS. SECTION  8. 

1:  A  farmer  wishes  to  mix  116  bushels  of  provender, 
consisting  of  rye,  barley,  and  oats,  so  that  it  may  contain  f 
as  much  barley  as  oats,  and  5  as  much  rye  as  barley.  How 
.  much  of  each  must  there  be  in  the  mixture  ? 

Stating  the  question,     a;=oats ;  and  *y=  barley. 
Then,^of^=f,=  rye. 

Forming  the  equation,  a;+*7-+Y|=116. 

Multiplying  by  14,  14a;+10a;+5a;=1624 

Uniting  terms,  29a;=1624 

Dividing  by  29,  a:=56     the  Ans. 

2.  I  paid  away  a  fourth  of  my  money,  and  then  a  fifth 
of  the  remainder,  which  was  $72.  How  much  money  had 
I  at  first  ? 


DIVISION  OF  FRACTIONS. 

» 

Stating  the  question. 

ic=  what  I  have. 
£^  paid  first. 
X — J=  the  remainder. 

Forming  the  equation. 
Multiplying  by  20, 
Uniting  terms, 
Dividing  by  3, 

1 — ^^=  paid  afterwards. 
l-w-72 
4a?— a:=1440 
3a?=1440 
a?=480    the  Ans. 

67 


3.  After  paying  away  ^  of  my  money,  and  then  |  of  the 
remainder,  I  had  872  left.     How  much  had  I  at  first  ? 

In  stating  the  question,  the  remainder  after  the  first  pay- 
ment was^  ;  and  \  of  that  is|J. 

Forming  the  equation,  x — ~ — 1^==72 

Multiplying  by  20,  20a?— 5a?— 3iP=1440 

Uniting  terms,  and  dividing,  a?=120 

4.  A  clerk  spends  f  of  his  salary  for  his  board,  and  |  of 
the  remainder  in  clothes,  and  yet  saves  $150  a  year.  What 
is  his  yearly  salary?  Ans.  81350. 

5.  Of  a  detachment  of  soldiers,  f  are  on  actual  duty,  | 
sick,  I  of  the  remainder  absent  on  leave,  and  there  are  380 
officers.     What  is  the  number  of  men  in  the  detachment? 

Ans.  1000  men. 

6.  A  young  man,  who  had  just  received  a  fortune,  spent 
1^  of  it  the  first  year,  and  J  of  the  remainder  the  next  year ; 
when  he  had  81420  left.     What  was  his  fortune? 

Ans.  811360. 

7.  If  from  ^  of  my  height  in  inches,  12  be  subtracted,  ^ 
of  the  remainder  will  be  2.     What  is  my  height  ? 

Ans.  5  ft.  6  in. 

8.  A  bowl  of  punch  was  mixed  as  follows  :  \  was  rum. 


68  ALGEBRA.  [Eq.  SeC.  8. 

c 

i  brandy,  ^j  acid  and  sugar,  and  3  pints  more  than  half  of 
all  these  was  water.     How  much  did  the  bowl  contain  ? 

Ans.  6  quarts. 

9.  A,  B,  and  C,  own  together  a  field  of  36  acres.  B  has 
•J  more  than  A,  and  C  has  i  more  than  B.  What  is  each 
man's  share?  Ans.  A,  9 acres;  B,  12  ;  C,  15. 

10.  A  gentleman  leaves  8315  to  be  divided  among  four 
servants  in  the  following  manner.  B  is  to  receive  as  much 
as  A,  and  |  as  much  more  ;  C  is  to  receive  as  much  as  A 
and  B,  and  ^  as  much  more  ;  D  is  to  receive  as  much  as  the 
other  three,  and  i  as  much  more.  What  is  ihe  share  of 
each?  Ans.  A  $24 ;  B  $36  :  C  $80  ;  D  $175. 


SUBTRACTION  OF  COMPOUND  QUANTITIES. 

§97.  Suppose  we  wish  to  subtract  the  expression  x+6 
from  y.  It  is  evident  that  we  may  first  subtract  x ;  which 
will  give  us,  y — x.  But  we  wish  to  subtract  not  only  x,  but 
6  also.  W^ell,  after  we  have  subtracted  x  we  will  subtract 
6  also ;  and  then  the  answer  will  be,  y — x — 6.  Therefore, 
whenever  we  wish  to  subtract  a  compound  quantity  whose 
terms  are  all  positive,  we  write  them  after  the  other  quan- 
tity with  -f  changed  to  — . 

§98.  Again,  suppose  we  wish  to  subtract  the  expression 
X — 6  from  y;  in  which  the  number  to  be  subtracted  has  — 
instead  of  -{-  •  As  before,  we  will  first  subtract  x,  by  which 
we  obtain  y — x.  But  the  quantity  to  be  subtracted  was  6 
less  than  x ;  and  we  have  therefore  subtracted  6  too  much. 
We  will  therefore  add  6  to  our  last  answer  for  the  true  re- 
mainder, which  will  give  us  y — a?-J-6.  Here,  we  have 
changed  the  positive  a:  to  —  ;  and  the  — 6,  to  -f-6. 


SUBTRA-CTION  OF  COMPOUND  QUANTITIES.  69 

§99.  Hence  the  propriety  of  the  following  rule  for  sub- 
tracting compound  quantities.  Change  all  the  signs  of  the 
expression  which  is  to  be  subtracted,  the  sign  -}-  to  — ,  and 
the  sign  —  to  -{■ ;  and  then  write  the  terms  after  the  other 
quantity.  It  is  to  be  recollected  that  in  the  quantityyrowi 
which  we  subtract,  the  signs  are  not  altered. 

§100.  Although  the  subtraction  is  performed  the  moment 
the  quantities  are  written  according  to  the  above  rule,  yet 
it  is  always  expedient  to  unite  the  terms  if  necessary  after 
the  operation. 

EXAMPLES. 

1.  Subtract  7a?+6+y  from  63/— 17. 

Ans.  6y — 17 — 7x — 6 — y;  which  =5y — 23 — 7x. 

2.  Subtract  4y+3a:— 10  from  74— ar. 

Ans.  74 — X — 42/ — 3a:+10;  which  =84 — 4ar — 4y. 

3.  Subtract  6 — x — Sy  from  7ic+6y. 

Ans.  7«+6y — d-^x-^Sy;  which  equals  8a; -|-9y — 6. 

4.  From  4a?— 3i/+27,  subtract  6y  — 12+ar. 

Ans.  3a; — 93/+ 39. 

5.  From  6-\-x — y,  subtract  13 — 9y — x. 

Ans.  2a;+82/— 7. 

6.  From  8a;— 2  +  3i/,  subtract  Sy-\-^—3x. 

Ans.  11a; — 6. 

7.  From  5+4a;,  subtract  2 — 5x-\-4:y — z. 

Ans.  9a; — 4y-f-3-f  2;. 

8.  From  6x — 8y,  subtract  — x — y-^50, 

Ans.  7a; — 7y — 50. 

2x 

9.  From  14 — ;^,  subtract  3a; — 12. 

o 

2r 

Ans.26— 3a;— -. 


70  ALGEBRA.  [Eq,.  SeC.  9. 


10.  From  — ^ — ,  subtract  5  +  —  . 
S  5 


,       8+7x     ^     Sx 

AnS.  :; 5 - 

3  5 


11.  From  7 —  -^,  subtract — ^- — 

§101.  It  has  been  shown,  §65  and  §66,  that  any  com- 
pound quantity  may  be  considered  and  operated  upon  as  a 
simple  quantity,  by  merely  drawing  a  vinculum  above  it,  or 
enclosing  it  in  a  parenthesis.  Whenever  that  compound 
quantity  is  a  fraction,  the  line  between  the  numerator  and 
denominator  serves  as  a  vinculum.     Thus,  in  the  eleventh 

example,  above,  — ^ is  subtracted  as  a  simple  quantity ; 

and  therefore  the  sign  in  it  is  not  changed. 

4a?— 6     Sx+7 


,„    ^        4.r— 6       ^         ^x+7 
12.  From  — : — ,  subtract  — -- 


,_    ^        3a;+8      ,  51 — X 

13.  rrom -—,  subtract  — -— . 


Ans. 


3a?+8     51— a; 
Ans.  — -: --. 


,.    ^        7— 2a;      ,  21x— 4 

14.  J:<rom — -,  subtract -— x. 

o  10 

„     2x    21a;— 4  , 

EOLATIONS. SECTION  9. 

1.  There  are  two  numbers,  whose  sum  is  140;  and  if  4 
times  the  less  be  subtracted  from  3  times  the  greater,  the 
remainder  is  70.     What  are  the  numbers  ? 


SUBTRACTION  OP  COMPOUND  QUANTITIES.  71 

Stating  the  question,  a?=  the  greater. 

140 — a?=  the  less. 
Forming  the  equations,       Sx — 560+4a?=70. 
Transposing  and  uniting,  07=90. 

Ans.  Greater  90  ;  Less  50. 

2.  A  person,  after  spending  $100  more  than  a  third  of 
his  yearly  income,  found  he  had  $150  more  than  half  of  it 
renaaining.     What  was  his  income  ?  Ans.  $1500. 

3.  Divide  the  number  48  into  two  such  parts,  that  the 
excess  of  one  of  them  above  20,  shall  be  three  times  as 
much  as  the  other  wants  of  20.  Ans.  32  and  16. 

4.  Two  men,  A  and  B,  commenced  trade.  A  had  twice 
as  much  money  as  B ;  he  has  since  gained  $50,  and  B  has 
lost  $90 ;  and  now  the  difference  between  A's  and  B^s  mo- 
ney, is  equal  to  three  times  what  B  has.  How  much  had 
each  when  they  commenced  trade  ? 

Ans.  A  $205;  B  $410. 

5.  A  gentleman  bought  a  watch  and  chain  for  $160.  If 
J  of  the  price  of  the  watch  be  subtracted  from  six  times  the 
price  of  the  chain,  the  remainder  will  be  the  same  as  if -/^ 
of  the  price  of  the  chain  were  subtracted  from  twice  the 
price  of  the  watch.     What  was  the  price  of  each? 

Ans.  Watch  $112  ;  Chain  $48. 

6.  Divide  the  number  204  into  two  such  parts,  that  if  f 
of  the  less  were  subtracted  from  the  greater,  the  remainder 
will  be  equal  to  ^  of  the  greater  subtracted  from  four  times 
the  less.  Ans.  Greater  154  ;  Less,  50. 

7.  Two  travellers,  A  and  B,  found  a  purse  of  money.  A 
first  takes  out  $2  arid  J  of  what  remains ;  and  then  B  takes 
out  $3  and  ^  of  what  remains  ;  and  it  is  found  that  each 
have  the  same  sum.     How  much  money  was  in  the  purse  ? 

Ans.  $20. 


72  ALGEBRA.  [Eq.  SeC.  9. 

8.  A  man  bought  a  horse  and  chaise  for  8341.  If  |  of 
the  price  of  the  horse  be  subtracted  from  twice  the  price  of 
the  chaise,  the  remainder  will  be  the  same  as  if  -f  of  the 
price  of  the  chaise  be  subtracted  from  three  times  the  price 
of  the  horse.     What  was  the  price  of  each  ? 

0:^  If  the  chaise  be  x,  and  the  horse  be  341 — x ;  then 

the  first  reniainder  will  be  2x — '— .     But  when  the 

8 

fraction  is  destroyed,  the  vinculum  is  taken  away  and  there- 
fore the  last  sign  must  be  changed  from  —  to  -+-. 

Ans.  Chaise  $189 ;  Horse  $152. 

9.  A  person  in  play  lost  a  fourth  of  his  money,  and  then 
won  back  3s.  ;  after  which  he  lost  a  third  of  what  he  now 
had,  and  then  won  back  2s. ;  lastly  he  lost  a  seventh  of 
what  he  then  had,  and  then  found  he  had  but  12s.  remain- 
ing.    What  had  he  at  first?  Ans.  205. 

10.  A  lady  spent  ^  of  her  money  at  the  linen  drapers,  |- 
of  the  remainder  at  the  mercers,  \  of  what  she  had  left  at 
the  milliners,  paid  3s.  for  a  coach,  and  carried  home  -^-^  of 
the  sum  she  had  at  first.     How  much  had  she  at  first  ? 

Ans.  210  shillings. 


UNITING  FRACTIONS  OF  DIFFERENT  DENOMINA- 
TORS. 

§102.  By  looking  at  the  answers  to  the  lasH  three  exam- 
ples in  §100,  and  also  the  three  in  §101,  it  will  appear  that 
we  ought  to  have  some  rule  for  uniting  their  terms.  We 
can  easily  find  one  by  applying  the  principle  explained  in 
§81.  For  we  have  only  to  change  each  of  the  fractions  to 
such  equivalent  fractions  as  will  have  one  with  another  the 
same  denoipinator. 


DIFFERENT  DENOMINATOHS.  W 

Thus,  the  answer  to  the  twelfth  example  under  §100,  is 

* Now  each  of  these  two  fractions  may 

4  5 

be  changed  to  20ths,  by  multiplying  the  first  by  5,  and  the 

20a;— 30       12a:+28 

""20  20 

(20x—'S0)—(Ux+2S)  _  20a?— 80— 12x— 28 

20  20 

Sx— 58 


last  by  4.     They  will  then  become 
which 


20 
§103.  Thus  we  have  the  rule  for  uniting  fractions  of 
different  denominators.  Multiply  all  the  denominators  to- 
gether for  a  new  denominator ;  and  each  numerator  by 
all  the  denominators  except  its  own^  for  new  numerators  ; 
remembering  that  if  a  compound  numerator  follows  — ,  all 
the  signs  in  it  must  be  changed  the  moment  one  denomina- 
tor is  used  for  the  whole  quantity, 

EXAMPLES. 

1.  Unite  the  terms  in  the  answer  to  the  13th  sum  in  §101. 
Operation. 
•Sx  +  S       51— a;_  9a:-f 24       102— 2a;_ 
~2  3  6  6 

9xrf24— 102  +  2x_  liar— 78 
6  —       6         * 

Note. — In  this  operation  the  first  minus  has  reference  to 

the  whole  quantity  V. i^^'y    the  second    minus  to  the 

o 

\  02 2x 

whole  quantity  ( — \    In  this  last  quantity,  102  has 

no  sign  before  i^seZ/*,  and  is  therefore  positive.  Now,  when 
the  line  between  the  numerator  and  denominator  is  carried 
through  the  whole  quantity,  the  vinculum  of  102 — 2x  is 

7 


74  ALGEBRA.  [Eq.  SeC.  9. 

destroyed ;  and  then  is  the  time  for  changing  the  signs  for 
subtracting. 

2.  Unite  the  terms  in  the  quantity  7 ^ ^-— -. 

o  o 

7 
Operation,    Here  we  have  7  which  =  — ,  so  tjjat  the 

.      .     ^                   7        2y       lOv— 4        „    ^ 
quantity  IS  the  same  as ^ ■.      Both  terms 

of  the  fraction  —  muhiplied  by  3  times  6=  -— -.         -^  [ 

1  lo  C    J 

_12y  10^-4)  30y-12 

,^^7        2y      10y-4_126       12y       30y~]2       ^^^.^^^ 
"13  6  18         18  18     ' 

126— 12y— 301/+12        u-  u  _  1>^8— 42y  _  2:3--7y 

=  Jo  ;  which  ^g     -      ^     . 

8+7x  3a; 

3.  Unite  the  terms  in  the  quantity,  — 5  — —  . 

40+35a?— 75— 9x      26a?— 35 
^°'- 15 =="^[5~' 

2x 

4.  Unite  the  terms  in  the  quantity,  26 — 3a? . 

78— 9a;— 2a:      78— 7a;      78      7a;      ^^      7ar 
Ans. -^   =._.^_=26--. 

TT  .       ,                .      ,              .      ^      52a:      21a;— 4 
J  5.  Unite  the  terms  m  the  quantity,  7 jj: — 


+a;. 

Ans.  — 

2— 53a; 
30       * 

6.  Unite  the  terms  in  the  quantity,  12- 

6+2a; 
4 

4a;— 3 
5 

12a; 

8  ' 

Ans 

111— 28ar 

10     mi  i^ii 

•  ,'.  Si^ifies  therefore. 

r,  '-:    ■;  ;■,' 

*■  -ii|U«i^.A'. 

DIFFERENT  DNEOMINATORS.  75 

X  X  X 

7.  Unite  the  terms  in  the  quantity,  J  +  3"  +  4"' 

13a;  X 

Afls.   —  =x+-. 

X  2x 

S.  Unite  the  terms  in  the  quantity,  4-1-  17  —  ^+"5  * 

Ans.l-h|. 

Sx  4z 

9.  Unite  the  terms  in  the  quantity,  42-}-  —  -{-z  —  — . 

7  5 

Ans.5.--. 

10.  Unite  the  terms  in   the  quantity,  y ^— — {-  2y— 

o 

3     * 

26i/— 3  ,    lly— 8 

Ans ^— — -  =  1/H —  . 

15         ^^        15 

§104.  Whenever  there  are  integers  to  be  united  with 
fractions,  they  may  be  changed  to  fractions,  by  putting  the 
number  1  under  them  for  the  denominator.   See  example  2. 

Thus,  6=  ^  ;     8=  -. 


76  0it*/m^LGEBTLA. 


RATIO  AND  PROPORTION. 


§105.  When  an  unknown  quantity  is  not,  either  by  it- 
self, or  in  some  connexion  with  others,  known  to  be  equal 
to  some  known  quantity  or  set  of  quantities ;  we  may  some- 
times find  that  there  is  a  comparison  between  it  and  some 
known  quantity,  which  is  the  same  as  the  comparison  be- 
tween two  known  quantities. 

Thus,  suppose  I  buy  27  yards  of  cloth  for  S72,  and  wish 
to  sell  for  816  so  much  of  it  as  cost  me  $16.  In  this  case 
the  number  of  yards  to  be  sold  is  not  equal  to  any  other 
quantity  that  is  mentioned.  But  we  suppose  that  it  must 
compare  with  the  number  of  yards  bought,  in  the  same 
manner  that  $16  compares  with  $72.  By  knowing  this 
comparison  we  can  find  the  number  of  yards ;  because,  as 
$16  is  |-  of  $72,  so  the  number  of  yards  to  be  sold  must 
be  I  of  the  number  of  yards  bought.     It  is  6  yards. 

§106.  It  will  be  seen  that  the  comparison  in  this  exam- 
ple, consists  in  observing  how  many  times  one  of  the  num- 
bers is  contained  in  the  other.  16  is  contained  in  72,  two 
ninths  of  a  time.  When  a  comparison  of  this  kind  is 
made,  the  result  that  is  obtained  is  called  their  ratio.  Thus, 
in  comparing  the  numbers  3  and  4,  we  find  that  4  is  con- 
tained in  3,  three-fourths  of  a  time ;  and  therefore  we  say 
the  ratio  ofSto^is^. 

§107.  The  pupil  must  remember  that  the  ratio  of  one 
number  to  another,  always  signifies  how  the frst  number 
compares  with  the  last.  Thus,  the  ratio  of  8  to  6,  is  | ; 
that  is,  8  is  I  of  5.  Hence  the  ratio  is  expressed  by  mak- 
ing the  Jirst  term  to  be  a  numtrator,  and  the  last  to  be  the 
denominator. 


RATIO  AND  PROPORTION.  T7 

§108.  In  the  example  just  furnished  relative  to  the  cloth ; 
the  ratio  of  the  money  paid,  to  the  money  obtained  for  a 
part  of  the  cloth  ;  (that  is,  the  ratio  of  $72  to  $16,)  is  ^|= 
|.  And  so  also  the  ratio  of  the  cloth  bought,  to  the  cloth 
sold  ;  (that  is,  the  ratio  of  27  yards  to  6  yards,)  is  y  which 
equals  |.  Here  we  see,  that  although  the  ratios  are  differ- 
ently expressed,  they  are,  notwithstanding,  equal  to  one 
another. 

§109.  When  the  ratio  of  two  quantities  is  equal  to  the 
ratio  of  ether  two  quantities,  there  is  said  to  be  a  propor- 
tion between  them  ;  that  is,  an  equality  of  ratios  is  called 
a  proportion, 

§110.  Our  chief  business  with  ratios  at  present,  is  to 
learn  when  they  form  a  proportion ;  that  is,  when  they 
are  equal  to  one  another.  Now,  as  they  may  be  express- 
ed in  the  form  of  a  fraction,  it  is  evident,  that  when  they 
are  brought  to  a  common  denominator,  if  they  are  propor- 
tional their  numerators  will  be  the  same,  and  if  they  are 
not  proportional  their  numerators  will  not  be  the  same. 

For 'example,  is  11  to  21=33  to  63?  We  pursue  our 
inquiry  as  follows :  11  to  21  is  the  same  as  |-J-,  and  33  to 
63  is  the  same  as  -^|.  We  bring  the  fractions  to  a  common 
denominator  by  §103. 

11       J  23        693         J  693 
2l""^  63  =1323  ^"^1323- 
We  find  they  are  equal,  and  the  four  terms  11  to  21ss33 
to  63  are  proportional. 

§111.  Although  ratios  are  sometimes  expressed  fraction- 
ally, they  are  generally  expressed  as  follows  ;  11  :  21  and 
33  :  63  ;  that  is  11  divided  by  21,  33  divided  by  63.  The 
pupil  w  ill  see  that  we  employ  the  same  sign  that  expresses 

7* 


division,  with  the  exception  of  the  —  between  the  two  dots. 
The  sign  :  is  read  is  to,  and  the  foregoing  examples  are 
read  11  is  to  21  and  33  is  to  63. 

§112.  When  four  quantities  are  proportional,  they  are 
written  thus,  11  :  21  :  :  33  :  63.  The  sign  : :  is  read  as  ; 
and  the  whole  expression  is  read,  11  is  to  21  as  33  is  to  63. 

§113.  In  a  proportion,  the  first  and  the  last  terms  are 
called  extremesyand  the  two  middle  terms  are  called  means. 
In  the  above  proportion,  11  and  63  are  the  extremes,  and 
21  and  33  are  the  means. 

^  §114.  In  order  to  derive  any  important  use  from  a  pro- 
portion, we  wish  the  pupil  to  recollect  the  method  employed 
td  find  whether  four  quantities  are  proportional.  We  mul- 
tiplied, (see  §110,)  the  first  numerator  by  the  last  denomi- 
nator,, to  find  one  new  numerator.  These  were  the  two 
extremes.  We  also  multiplied  the  last  numerator  by  the 
first  denominator,  to  find  the  other  new  numerator.  These 
were  the  two  means.  And  hence  we  learn,  that  if  four 
quantities  are  proportional,  the  product  of  the  two  extremes 
is  equal  to  the  product  of  the  means. 

§115.  Therefore,  whenever  in  our  operations  we  find  a 
proportion,  we  can  easily  reduce  it  to  an  equation  by  mul- 
tiplying the  extremes  together  for  one  member ;  and  multi- 
plying the  means  together  for  the  other  member.  Thus, 
2  :  7  : :  8  :  a:,  becomes  in  an  equation  2a7=56  ;  whence 
:r=28.  ■ .  •'^  wm  ^  im»- ^^ 

-f?  f-^i,>ry^    BaUATIONS. SECTION  10.  Jiff 

1.  Tf  you  divide  $75  between  two  men  in  the  proportion 
of  8  to  2,  what  will  each  man  receive  ? 


RATIO  AND  PROPORTION.  79 

Stating  the  question,  x=  tlie  share  of  one. 

75 — a?=  the  share  of  the  other. 
Making  the  proportion,  x ;  75 — x  : :  3  :  2 

Reducing  to  an  equation,  2x=225 — 3a; 

Transposing  and  uniting,  5a:=225 

Dividing,  a?=45 

Ans.  $45  ;  and  $30. 

2.  Divide  150  into  two  parts,  so  that  the  parts  may  be 
to  each  other  as  7  to  8.  Ans.  70 ;  and  80. 

3.  Divide  $1235  between  A  and  B,  so  that  A's  share 

may  be  to  B's  as  3  to  2. 

Ans.  A's  share  $741 ;  B's  $494. 

4.  Two  persons  buy  a  ship  for  $8640.  Now  the  sum 
paid  by  A  is  to  that  paid  by  B,  as  9  to  7.  What  sum  did 
each  contribute?  Ans.  A  paid  $4860  ;  B  $3780. 

5.  A  prize  of  $2000  was  divided  between  two  persons, 
whose  shares  were  in  proportion  as  7  to  9.  What  was  the 
share  of  each?  Ans.  $875  ;  and  $1125. 

6.  A  gentleman  is  now  30  years  old,  and  his  youngest 
brother  20.  In  how  many  years  will  their  ages  be  as  5  to 
4? 

|C?^  After  stating  the  question  the  proportion  will  be 
30 +a;:   20  +  0" : :  5:4.  Ans.  20  years. 

7.  What  number  is  that,  which,  when  added  to  24,  and 
also  to  36,  will  produce  sums  that  will  be  to  each  other  as  7 
to  9?  Ans.  18. 

8.  Two  men  commenced  trade  together.  The  first  put 
in  $40  more  than  the  second ;  and  the  stock  of  the  first 
was  to  that  of  the  second  as  5  to  4.  What  was  the  stock 
of  each?  Ans.  $200;  and  $160. 

9.  A  gentleman  hired  a  servant  for  $100  a  year,  toge- 
ther with  a  suit  of  clothes  which  he  was  to  have  immedi- 


80  ALGEBRA.  [Eq.  SkC  11. 

ately.  At  the  end  of  8  months,  the  servant  went  away, 
and  received  $60  and  kept  the  suit  of  clothes.  What  was 
the  value  of  the  suit  of  clothes  ?  Ans.  $20. 

10.  A  ship  and  a  boat  are  descending  a  river  at  the  same^ 
time ;  and  when  the  ship  is  opposite  a  certain  fort,  the  boat 
is  13  miles  ahead.    The  ship  is  sailing  at  the  rate  of  5  miles 
while  the  boat  is  going  3.    At  what  distance  below  the  fort, 
will  they  be  together?  Ans.  32^  miles. 

§116.  It  is  very  often  the  case  that  a  problem  is  easily 
solved  by  using  simply  the  ratioj  instead  of  a  proportion. 

.      ';        EQUATIONS. SECTION  11. 

Operation  by  Ratio. 

.1.  Divide  40  apples  between  two  boys  in  the  proportion 

of  3  to  2. 

Stating  the  question,  x=  the  share  of  one. 

Now,  as  the  ratio  of  the  first  to  the  second  is  J  ;  then  the 

ratio  of  the  second  to  the  fii-st  is  |.     Therefore, 

2ic 

---  =  the  share  of  the  second. 

3 

2x 
Forming  the  equation^    ^+-^  =40 

o 

Multiplying  by  3,  3a?4-2x=  120 

Consequently,  a?=  24. 

Ans.  24;  and  16. 
2.  Three  men  trading  in  company,  gain  $780.    As  often 
as  A  put  in  $2,  B  put  in  $3,  and  C  put  in  $5.     What  part 
of  the  gain  must  each  of  them  receive  ] 


Stating  the  question,  a;=  A's  share. 

— -  ■=  B's  share.  ''  > 

-~-  =.  C's  share^  <  • 


•..  .* 


RATIO  AND  PROPOKTION.  81 

Forming  the  equation,       ^+-77-  -\ — 17  =  '^^^' 

lit  4t 

Ans.  A  $156;  B  $234  ;  C$390. 

3.  Two  butchers  bought  a  calf  for  40  shillings,  of  which 
the  part  paid  by  A,  was  to  the  part  paid  by  B,  as  3  to  5* 
What  sum  did  each  pay?  Ans.  A  pa'd  15s. ;  B  25s. 

4.  Divide  560  into  two  such  parts,  that  one  part  may  be 
to  the  other  as  5  to  2.  Ans.  400 ;  and  160. 

5.  A  field  of  864  acres  is  to  be  divided  among  three  far- 
mers, A,  B,  and  C  ;  so  that  A's  part  shall  be  to  B's  as  5  to 
11,  and  C  may  receive  as  much  as  A  and  B  together.  How 
much  must  each  receive? 

Ans.  A  135  ;  B  297  ;  C  432  acres. 

6.  Three  men  trading  in  company,  put  in  money  in  the 
following  proportion  :  the  first  3  dollars  as  often  as  the  se- 
cond 7,  and  the  third  5.  They  gain  $960.  What  is  each 
man's  share  of  the  gain?  Ans.  $192  ;  $448  ;,  $320. 

7.  Find  two  numbers  in  the  proportion  of  2  to  1,  so  that 
if  4  be  added  to  each,  the  two  sums  will  be  in  proportion  of 
3  to  2.  Ans.  8  and  4. 

8.  Two  numbers  are  to  each  other  as  2  to  3 ;  but  if  50 
be  subtracted  from  each,  one  will  be  one  half  of  the  other. 
What  are  the  numbers?  Ans.  100  and  150. 

9.  A  sum  of  money  is  to  be  divided  between  two  persons, 
A  and  B ;  so  that  as  often  as  A  takes  $9,  B  takes  $4. 
Now  it  happens  that  A  receives  $15  more  than  B.  What 
is  the  share  of  each  ?  Ans.  A  $27  ;  B  $12. 

10.  There  are  two  numbers  in  proportion  of  3  to  4  ;  but 
if  24  be  added  to  each  of  them,  the  two  sums  will  be  in  the 
proportion  of  4  to  5.     What  are  the  numbers  ? 

Ans.  72  and  96. 


9^  4. /.        •    ALGEBRA.  [Eq.  SeC.  11. 

11.  A  man's  age  when  he  was  married  was  to  that  of 
his  wife  as  a  to  2 ;  and  when  they  had  Hved  together  4 
years,  his  age  was  to  hers  as  7  to  5.  What  were  their 
ages  when  they  were  married  ? 

Ans.  His  age  24;  hers  16  years. 

12.  A  certarn  man,  found  when  he  married,  that  his  age 
was  to  that  of  his  wife  as  7  to  5.  If  they  had  been  married 
8  years  sooner,  his  age  would  have  been  to  hers  as  3  to  2. 
What  were  their  ages  at  the  time  of  their  marriage  ? 

Ans.  His  age  56  years  ;  hers,  40. 

13.  A  man's  age,  when  he  was  married,  was  to  that  of 
his  wife  as  6  to  5  ;  and  after  they  had  been  married  8  years, 
her  age  was  to  his  as  7  to  8.  What  were  their  ages  when 
they  were  married  ?  Ans.  Man  24  ;  Wife  20  years. 

14.  A  bankrupt  leaves  $8400  to  be  divided  among  four 
creditors,  A,  B,  C,  and  D,  in  proportion  to  their  claims. 
Now,  A's  claim  is  to  B's  as  2  to  3  ;  B's  claim  to  C's  as  4 
10  5  ;  and  C's  claim  to  D's  as  6  to  7.  How  much  must 
each  creditor  receive? 

Ans.  A  81280  ;  B  $1920  ;  C  $2400  ;  D  $2800. 

15.  A  sum  of  money  was  divided  between  two  persons, 
A  and  B,  so  that  the  share  of  A  was  to  that  of  B  as  5  to  3. 
Now,  A's  share  exceeded  J-  of  the  whole  sum  by  $50. 
What  was  the  share  of  each  ?  Ans.  $450  ;  and  $270. 


EaUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  83 


EQUATIONS  WITH  TWO  UNKNOWN   QUANTITIES. 

§117.  It  frequently  happens,  that  several  unknown  quan- 
tities are  introduced  into  a  problem.  But  when  this  is  the 
case,  if  the  conditions  will  give  rise  to  as  many  equations, 
independent  of  each  other,  as  there  are  unknown  quanti- 
ties, there  is  no  difficulty  in  finding  the  value  of  each  quan- 
tity. 

§118.  An  equation  is  said  to  be  independent  of  another 
when  it  cannot  be  changed  into  that  other.  Thus,  7x — y 
=47,  is  independent  of  the  equation  10y-^4x=50;  be- 
cause one  of  them  cannot  be  so  altered  ns  to  make  the  other. 
But,  7x — 1/=47,  is  not  independent  of  the  equation  21  ar — 
8^=141 ;  because  the  last  is  made  by  multiplying  the  first 
by  3. 

§119.  At  present  we  will  attend  only  to  equations  that 
include  two  unknown  quantities,  each  represented  by  a  dif' 
ferent  letter  from  the  other. 

§120.  In  equations  that  contain  two  unknown  quanti- 
ties, our  first  object  must  be  to  find  the  value  of  one  of 
them ;  and  in  order  to  do  this,  the  preliminary  step  is  to 
derive  from  the  equations  that  are  given,  another  equation 
which  shall  have  but  one  unknown  quantity.  This  opera- 
tion is  called  exterminating  the  other  unknown  quantities. 

§121.  There  are  three  different  methods  of  forming  one 
equation  with  one  unknown  quantity  from  two  equations 
containing  two  unknown  quantities.  With  each  of  these, 
the  learner  should  become  familiar;  as  it  is  sometimes  con- 
venient to  use  one  of  them,  and  sometimes  another. 


84  ,^r^,T^I^^.^,      ALGEBRA.       iTr^HH^I**  ^6^« 

First  Method. 

§122.  It  is  necessary  here  to  recollect  what  was  stated  in 
§42,  that  when  equals  are  added  to  equals,  their  sums  will 
be  equal ;  and  also  when  equals  are  subtracted  from  equals, 
their  remainders  are  equal.  Thus,  suppose  we  have  the 
equation,  a?+14=36  ;  and  suppose  also  that  we  know  that 
y=8  ;  ihen  if  we  will  add  y  to  the  first  member,  and  8  to 
the  last,  the  members  will  still  be  equal  to  one  another,  as 
follows:  x+l4.-\-y—S6+8.  And  also  if  we  subtract  y 
from  the  first  member,  and  8  from  the  last,  the  members 
will  still  be  equal  to  one  another;  thus  aj-f-14 — ^2/==^ — 8» 

§12'3.  This  principle  can  be  easily  applied  for  the  exter- 
mination of.  unknown  quantities.  For,  if  in  both  of  two 
equations,  one  of  the  unknown  quantities  has  the  same  co- 
effccieni,  but  after  different  signs,  it  is  evident  that  if  we 
add  both  equations  together,  viz.  the  first  member  to  the  first 
member,  and  the  last  member  to  the  last  member ;  and 
then  unite  terms,  a  new  equation  will  be  formed  in  which 
that  unknown  quantity  will  disappear. 

EXAMPLES. 

1.  Given  the  two  equations,  <  -  _.2iy=38  v  ^^  ^"^  ^^^ 
values  of  x  and  y. 

Add  together  the  two  right  hand  members,  and  also  the 
two  left ;  and  we  have  the  equation 

3a:  +  21/ +5x— 21/^26  +  38 
Uniting  terms,  8a:=64 

Dividing,  a:=8 

Now,  ifz^^^Xhen  3a:=24 ;  and  the  first  equation  will 
become  .-^'fij  iiVs  24+2y=26,  from  which  we  may 
find  the  value  of  y.     .1^*  W**  r*  f^^ii  W  tw»*  m**  vi  kwitn^t 


EQUA-TIONS  WITH  TWO  UNKNOWN  QUANTITIES.  85 

2.  Given  the  equations,  <  qSZ./Uj^28  i  ^°  ^^^  ^^® 
values  of  a:  and  y.  Ans.  a;=6  ;  i/=4. 

3.  Given  the  equations,  <  o-cXe  ---S4.  i  ^°  ^^^  ^^® 
values  of  ic  and  y* 

OCr  In  this  example,  it  is  plain  that  the  y^s  will  not  be 
destroyed  by  adding  them  together.  But  we  have  before 
seen,  §59,  that  if  all  the  signs  are  changed,  the  equation  will 
not  be  affected.  Let  us  then  change  the  signs  of  the 
second  equation.      The  two  equations  will  then  become 

< 'o^_«'^Z 34  (  which,  when  added  together,  become 

5x+6y--2a?— 6i/=58— 34,  or  3a?=24.  .•.a;=8;  and  the 
first  equation  becomes  40+6i/=58.    Whence,  y=S. 

§124.  In  the  last  example,  if  we  take  the  equations  before 
the  alteration  of  the  second,  thus,  <  o^iXfi  — S4  i  ^"^ 
subtract  the  second  from  the  first,  the  result  will  be  the  same 
as  it  was  by  changing  the  signs.  As  follows ;  5x-{-6y — 
2a;— 6i/=58— 34. 

Whence  we  learn  that,  if  in  both  equations  one  of  the 
unknown  quantities  has  the  same  co-efficient  and  also  the 
same  sign,  and  we  subtract  one  equation  from  the  other, 
(viz.  the  first  member  from  the  first  member,  and  the  se- 
cond member  from  the  second  member,)  and  unite  the 
terms ;  we  shall  form  a  new  equation  in  which  that  un- 
known quantity  will  disappear. 

EXAMPLES. 

4.  Given  the  equations,  j  g^tS^Sl  (  ^°  ^"^  ^^® 
values  of  ar  and  yl  Ans.  a;=5  ;  y=7. 

8 


86  ALGEBRA. 

^•^^^^^    I  2^-6^58  (    to  find  :r  and  y. 

Ans.  x=2;  y=9* 

6.  Given    ^  ^ZepIoG  |    to  find  a^  and  y. 

Ans.  a?=40 ;  2/=lI' 

7.  Given    J  }2^21j,=63  J    ^  A-d  ^  and  3,. 

Ans.  x=7 ;  y^^l- 

8.  Given    i  3^Z.2v=  7  (    ^°  ^"^^  ^  ^"^  2/* 

Ans.  ar=5;  y=4. 

9.  Given    ^  _4^]}]2y=— 12  |    to  find  a;  and  y. 

Ans.  x=4: ;  y=2. 

10.  Given  J^^+2y;j4j    to  find  ^  and  y. 

(jfO"  In  this  example,  neither  of  the  unknown  quantities 
has  the  same  co-efl5cient  in  both  equations.  But  both  mem- 
bers of  the  last  equation  can  be  multiplied  by  3,  without 
destroying  the  equality,  §77  ;  and  then  the  a;'s  will  be  alike 

in  both  equations.   Thus,  )  3^T6^Z42  ( 

Ans.  a?=10;  y=2. 

11.  Given  ^  3^:]:4p^88  ]    ^°  ^"^  ^  ^^^  ^' 
^d^  The  second  may  be  multiplied  by  2. 

Ans.  a;=8;  y=l6» 

12.  Given  ^  ^l^yZ^l  I    *°  ^"^  ^  ^"^  2/- 

OO'  Make  the  j/'s  alike.  Ans.  x=S ;  y=7, 

13.  Given  ^  4^;;j:3p22  ^    to  find  x  and  2^. 

OO"  Multiply  the  first  by  2  and  the  a;'s  will  be  alike. 

Ans.  a:=4;  y=2» 


EaUA-TIONS  WITH  TWO  UNKNOWN  QUANTITIES.         87 

'*-<^-^-\elXl:=ll\     to  find.  and.. 

Ans.  x=7 ;  z=8. 

15.  Given   J  ^^l|=Jo  |    to  find  :.  and  y. 

Ans.  07=4;  3/=5. 

16.  Given  |  4^jl3pi7  \    to  find  x  and  y. 

Ans.  a?=8;  2/=  5* 

17.  Given  {'^iXlZlf  I  '°  ^"^  ^  ^"^ '■ 

Ans.  y=24;  «=6. 

18.Given52^+%^=7^    to  find  .=  and  j,. 

Ans.  x=2;  y=l' 

19.  Given  lly+llZlll    to  find  ,  and  a.. 
Multiply  the  first  by  4,  20?/  +  12ar==372 
Multiply  the  second  by  3,  %+12a:=:240 

Subtracting  the  2d  from  the  1st,  lly        =132. 

Ans.  2/=12;  a;=ll. 

20.  Given  5  l^~llZ^  I    to  find  2j  and  ^. 

Multiply  the  first  by  5,  2O3/— 252=10 

Multiply  the  second  by  4,  20y— 16z=28 

Subtracting  the  1st  from  the  2d,  92=18. 

Ans.  2/=  3  ;  z=2. 

§125.  From  the  foregoing,  we  derive  the  following  : 

Rule  I.  to  exterminate  an  unknown  quantity. 

Determine  which  of  the  unknown  quantities  you  will  ex- 
terminate ;  and  then^  if  it  is  necessary,  multiply  or  divide 
one  or  both  of  the  equations  so  as  to  make  the  term  which 
contains  that  unknown  quantity  to  be  the  same  in  both. 

Then  if  the  identical  terms  have  like  signs  in  both 


88  ALGEBRA.  [Ea.  Sec.  12. 

equations,  subtract  one  equation  from  the  other ;  hut  if 
they  have  unlike  signs,  add  one  equation  to  the  other. 
And  the  result  will  be  an  equation  containing  only  one  un- 
known quantity.  .       , 

EQUATIONS. SECTION  12. 

1.  What  two  numbers  are  those  whose  sum  is  20  and 
difference  12  ?  » 

Stating  the  question,  x=  greater  number. 

y=  the  less. 
Then  forming  the  equations,  x+y=:20 

X — 2/ =12 
Adding  the  equations,  2a;    =32  .•.  a;=16 

Substituting  16  for  a;  in  the  first,     16+2/=20 
Transposing  and  uniting,  2/=  4. 

Ans.  16  and  4. 

2.  A  market  woman  sells  to  one  person,  3  quinces  and 
4  melons  for  25  cents ;  and  to  another,  4  quinces  and  2 
melons,  at  the  same  rate,  for  20  cents.  How  much  are  the 
quinces  and  melons  apiece? 

Forming  the  equations,  3ar-f  42/=25 

4a;+2i/=20 
Multiplying  the  second  by  2,  8a?+4i/=40 

Subtracting  1st  from  2d,  .5a?        =15. 

Ans.  Quinces  3  cents ;  Melons  4. 
In  any  of  our  solutions  after  this,  we  shall  number  the 
lines,  so  that  any  reference  to  them  will  be  easily  under- 
stood. 

3.  A  man  bought  3  bushels  of  wheat  and  5  bushels  of 
rye  for  38  shillings ;  and  at  another  time  6  bushels  of  wheat 
and  3  bushels  of  rye  for  48  shillings.  What  was  the  price 
for  a  bushel  of  each  ? 


EQUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  89 

Let     x=  price  of  wheat,  and  2/=  price  of  rye. 

1.  By  the  first  condition,  3a:4-5y=38 

2.  By  the  second,  6a;+3y=48 

3.  Multiplying  the  1st  by  2,         6a?+10//=76 

4.  Subtracting  the  2d  from  the  3d,  7i/=28  .-.i/=4 

5.  Substituting  4  for  yin  the  1st,  3x+20=38 

6.  Transposing  and  dividing,  37=6. 

Ans.  Wheat  for  6s. ;  Rye  for  4s. 

4.  Two  purses  together  contain  8400.  If  you  take  #40 
out  of  the  first  and  put  them  into  the  second,  then  there  is 
the  same  in  each.     How  many  dollars  does  each  contain  1 

Let     3?=  the  number  in  the  first. 
y  =  the  number  in  the  second. 

1.  By  the  first  condition,  a;+y=400 

2.  By  the  second,  x — 40=2/4-40 

3.  Transposing  the  2d,  x — y  ^80 

4.  Adding  the  1st  to  the  3d,  2x       =480 .-.  37=240. 

Ans.  The  first  $240  ;  the  second  $160. 

5.  A  gentleman  being  asked  the  age  of  his  two  sons,  re- 
plied that  if  to  the  sum  of  their  ages  25  be  added,  this  sum 
will  be  double  the  age  of  the  eldest ;  but  if  8  be  taken  from 
the  diflierence  of  their  ages,  the  remainder  will  be  the  age  of 
the  youngest.     What  is  the  age  of  each  1 

Let  x=  the  age  of  the  eldest,     y^  the  age  of  the  youngest. 

1.  By  the  first  condition,  x-\-y+2^=z2x 

2.  By  the  second,  x — y — 8  ==y 

3.  Transposing  and  uniting  1st,      — x-{-y~ — 25 

4.  Transposing  and  uniting  2d,         x — 2i/^  8 

5.  Adding  3d  and  4th,  —y=—n 

6.  Substituting  17  for  y  in  the  3d,  —37+17  =  —25 

7.  Transposing  — 37=  — 42 

Ans.  Eldest  42  ;  Youngest  17- 
8* 


90  ALGEBRA.  [Eq.  SeC.  12. 

6.  A  gentleman  paid  for  6  pair  of  boots  and  4  pair  of 
shoes  $44 ;  and  afterwards  for  3  pair  of  boots  and  7  pair 
of  shoes  832.     What  was  the  price  of  eacli  per  pair? 

Ans.  Boots  $6 ;  Shoes  $2. 

7.  A  man  spends  30  cents  for  apples  and  pears,  buying 
his  apples  at  the  rate  of  4  for  a  cent,  and  his  pears  at  the 
rate  of  5  for  a  cent.  He  afterwards  let  his  friend  have  half 
of  his  apples  and  one-third  of  his  pears  for  13  cents,  at  the 
same  rate.     How  many  did  he  buy  of  each  sort  ? 

Let    x=  number  of  apples. 
y=  number  of  pears. 

-J  cent  =  price  of  1  apple. 

—  cent  =  price  of  1  pear. 
5 


—  cents  =  price  of  all  the  apples. 
^  cents  =  price  of  all  the  pears. 


1.  By  the  first  condition,  4+  I"  =  30 

4      o 

2.  By  the  second,  5-f  X=  13 

o  lO 

8.  Dividing  the  1st  by  3,  ^^  -{-^  =  10 

12       15 

4.  Subtracting  3d  from  2d,        2.  —^^=  3 

8        12 

5.  Multiplying  by  24,  3a?— 2ar=:72  .-.  ir=72. 

Ans.  72  apples ;  60  pears. 

8.  One  day  a  gentleman  employs  4  men  and  8  boys  to 

labour  for  him,  and  pays  them  405. ;  the  next  day  he  hires 

at  the  same  rate,  7  men  and  6  boys,  for  505.     What  are 

the  daily  wages  of  each  1    Ans.  Man's,  4*. ;  boy's,  2s  Qd. 


EQUATIONS  WITH  TWO  UNKNOWN  QUANTITIES.  91 

9.  It  is  required  to  find  two  numbers  with  the  following 
properties.  ^  of  the  first  with  -^  of  the  second  shall  make 
16  :  and  ^  of  the  first  with  ^  of  the  second  shall  make  9. 

Ans.  12  and  30. 

10.  Says  A  to  B,  give  me  55.  of  your  money,  and  I  shall 
have  twice  as  much  as  you  will  have  left.  Says  B  to  A, 
give  me  55.  of  your  money,  and  I  shall  have  three  limes  as 
much  as  you  will  have  left.     What  had  each  ? 

Ans.  A  lis;  B  13s. 

11.  Two  men  agree  to  buy  a  house  for  $1200.  Says 
A  toB,  give  me  f  of  your  money,  and  I  shall  be  able  to  pay 
for  it  all ;  No,  says  B,  give  me  |  of  yours,  and  then  I  can 
pay  for  it.     How  much  money  had  each  ? 

Ans.  A  8800  ;  B  $600. 

12.  Find  two  numbers  with  the  following  properties. 
The  products  of  the  first  by  2,  and  the  second  by  5,  when 
added  are  equal  35.  Also,  the  products  of  the  first  by  7, 
and  the  second  by  4,  when  added  are  equal  to  68. 

Ans.  8  and  3. 

13.  A  paid  B  20  guineas,  and  then  B  had  twice  as  much 
money  as  A  had  left ;  but  if  B  had  paid  A  20  guineas,  A 
would  have  had  three  times  as  much  as  B  had  left.  What 
sum  did  each  possess  at  first  ? 

Ans.  A  52  guineas ;  B  44. 

14.  A  person  has  a  saddle  worth  £50,  and  two  horses. 
When  he  saddles  the  poorest  horse,  the  horse  and  saddle 
are  worth  twice  as  much  as  the  best  horse ;  but  when  he 
saddles  the  best,  he  with  the  saddle  is  worth  three  times  the 
poorest.     What  is  the  value  of  each  horse? 

Ans.  Best  £40 ;  Poorest  £30. 

15.  A  merchant  sold  a  yard  of  broadcloth  and  3  yards  of 
velvet  for  $25  ;  and,  at  another  time,  4  yards  of  broadcloth 


92  ALGEBRA.  [Eq.  SeC.  12. 

and  5  yards  of  velvet  for  $65.    What  was  the  price  of  each 
per  yard?  Ans.  Broadcloth  $10  ;  Velvet  $5. 

16.  A  person  has  500  coins  consisting  of  eagles  and 
dimes  ;  and  their  value  amounts  to  $1931.  How  many  has 
he  of  each  coin  ? 

^CT"  The  solution  must  be  in  cents. 

Ans.  190  eagles;  310 dimes. 

17.  In  the  year  1299,  three  fat  oxen  and  6  sheep  toge- 
ther cost  79  shillings  ;  and  the  price  of  an  ox  exceeded  the 
price  of  12  sheep  by  10  shillings.  What  was  the  value  of 
each?  Ans.  An  ox  24  shillings  ;  a  sheep  Is.  2d. 

18.  Two  persons  talking  of  their  ages^  A  says  to  B,  8 
years  ago  I  was  three  times  as  old  as  you  were ;  and  4 
years  hence,  I  shall  be  only  twice  as  old  as  you.  What 
are  their  present  ages  ?  Ans.  A  44 ;  B  20  years. 

19.  A  farmer  sold  to  one  man  30  bushels  of  wheat  and 
40  of  barley  for  270  shillings  ;  and  to  another,  50  bushels 
of  wheat  and  30  of  barley  for  340  shillings.  What  was 
the  price  per  bushel  of  each  ?   Ans.  Wheat  5s. ;  Barley  3s. 

20.  A  man  and  his  wife  and  child  dine  together  at  an 
inn.  The  landlord  charged  15  cents  for  the  child,  and  for 
the  woman  he  charged  as  much  as  for  the  child,  and  -^  as 
much  as  for  the  man ;  but  for  the  man  he  charged  as  much 
'as  for  the  woman  and  child  together.  What  did  he  charge 
for  each  ? 

Ans.  45  cents  for  the  man ;  and  30  cents  for  the  woman. 

21.  A  gentleman  has  two  horses,  and  also  a  chaise  worth 
^250.  If  the  first  horse  be  harnessed,  he  and  the  chaise 
will  be  worth  twice  as  much  as  the  second  horse ;  but  if  the 
second  be  harpessed,  he  and  the  chaise  will  be  worth  three 
times  as  much  as  the  first  horse.  What  is  the  value  of  each 
horse?  Ans.  First  $150;  second  $200. 


SECOND  METHOD  OP  EXTERMINATION.  93 

22.  A  is  in  debt  $1200,  and  Bowes  $2500;  but  nei- 
ther has  enough  to  pay  his  debts.  A  says  to  B,  lend  me 
the  I  of  your  fortune,  and  then  I  can  pay  my  debts.  But 
B  answered,  lend  me  the  |-  of  your  fortune,  and  I  can  pay 
my  debts.     What  was  the  fortune  of  each  ? 

Ans.  A  $900;  B  $2400. 

23.  A  wine  merchant  has  two  kinds  of  wine,  one  at  5s.  a 
gallon,  and  the  other  at  12s.;  of  which  he  wishes  to  make 
a  mixture  of  20  gallons  that  shall  be  worth  8s.  a  gallon. 
How  many  gallons  of  each  sort  must  he  use  1 

Ans.  8^  gallons  of  that  at  12s. ;  11^  of  that  at  5s. 


SECOND  METHOD  OF  EXTERMINATION. 

§126.  In  each  of  the  preceding  questions,  we  first  found 
the  value  of  owe  of  the  unknown  quantities  ;  and  then  sub- 
stituted that  value  for  that  unknown  quantity  in  one  of  the 
equations,  in  order  to  find  the  value  of  the  other  unknown 
quantity.  This  mode  of  operating  furnishes  a  hint  that 
leads  us  to  another  method  of  extermination. 

Let  us  take  the  first  question  in  the  last  section  [p.  88,3 

in  which  we  have  the  equations,  <        ^     io  i 

The  last  part  of  our  operation  was  to  substitute  the  value 
of  ar  for  x  itself,  in  one  of  the  equations.  It  is  evident  that 
we  could  make  this  substitution  just  as  well  if  the  value  of 
X  was  a  literal  quantity,  instead  of  16.     Thus,  supposing 


94  ALGEBRA.  [Eq.  SeC.   13. 

X  to  be  equal  to  ~ ;  then  substituting  it  for  a;,  the  first  equa- 
te 

tion  would  be  |-  +3/=  20. 

^.  t 

§127.  Let  us  therefore  transpose  the  first  equation  to  find 
what  X  will  equal,  just  as  if  we  knew  the  value  of  y.  We 
shall  find  that  a?^20 — ?/.  And  then  in  the  second  equation, 
we  will  use  the  value  of  a;  instead  of  x  itself. 

Thus,  20— y— 1/=12. 
Transposing  and  uniting,     — 2y=  — 8  .*.  y=  4 ; 
which  was  our  answer  by  the  first  method.     Then  x  will 
be  found  by  substituting  4  for  y.     Whence  we  derive 

RVLE  II.    TO  EXTERMINATE  AN  UNKNOWN  QUANTITY. 

§128.  By  one  of  the  equations,  find  the  value  of  one  of 
the  unknown  quantities,  as  if  the  other  were  known  ;  and 
then,  in  the  other  equation,  substitute  this  value  for  the  un- 
known quantity  itself. 

EQUATIONS. SECTION  13. 

1.  There  are  two  numbers  whose  sum  is  100  ;  and  three 
times  the  less  taken  from  twice  the  greater,  leaves  150  re- 
mainder.    What  are  those  numbers  ? 

Let     x=  greater. 

2/=  less. 
2x — Sy  -=  the  required  subtraction. 

1.  By  the  first  condition,  oc+y=  100 

2.  By  the  second,  2x — Sy=i60 

3.  Transposing  the  1st,  x=100—y 

4.  Multiplying  the  3d  by  2,  2a?     200— 2i/ 

5.  Substituting  200—2?/  for  x  in  the  2d,  '200— 2y— 3y=  150 

6.  Transposing  and  uniting,  — 5y  -=  — 50  .•.y=lO 


SECOND  METHOD  OF  EXTERMINATION.  95 

7.  Substituting  10  for  y  in  the  1st,     a? +10=  100 

8.  Transposing  and  uniting,  a? =90 

Ans.  Greater  90;  Less  10, 

2.  The  ages  of  a  father  and  his  son  amounted  to  140 
years  ;  and  the  age  of  the  father  was  to  the  age  of  the  son 
as  3  to  2.     What  were  their  ages  ? 

Let     a?=  age  of  the  father. 
y=  age  of  the  son. 

1.  By  the  first  condition,  37+3/=  140 

2.  By  the  second,  2/==  "^J 

o 

^3.  Transposing  the  1st,  2/=il40 — x 

2x 

4.  Substituting  140— x  in  the  2d,     l40-~a?=  — 

5.  Multiplying  by  3,  420— 3ar=2a? 

6.  Transposing  and  dividing,  ar=84 

7.  Substituting  84  for  x  in  3d,  2^=140—84=56. 

Ans.  Father  84  years  ;  Son  56. 

3.  Find  two  numbers,  such  that  |  of  the  first  and  i  of 
the  second  shall  be  87  ;  and  -J-  of  the  first  and  -J-  of  the  se- 
cond shall  be  55.  Ans.  135  ;  and  168. 

4.  A  says  to  B,  give  me  100  of  your  dollars,  and  I  shall 
have  as  much  as  you.  B  replies,  give  me  100  of  your  dol- 
lars and  I  shall  have  twice  as  much  as  you.  How  many 
dollars  has  each?  Ans.  A  $500 ;  B  700. 

5.  Two  servants  went  to  market.  A  laid  out  as  much 
above  4  shillings,  as  B  did  under  6  ;  and  the  sum  spent  by 
A  was  to  that  spent  by  B,  as  7  to  8.  How  much  did  each 
lay  out  ?  Ans.  A,  4s.  Sd, ;  B,  5«.  4d 

6.  Find  two  numbers  in  the  proportion  of  2  to  1 ,  so  that 
if  4  be  added  to  each,  their  two  sums  shall  be  in  proportion 
of  3  to  2.  Ans.  8  and  4. 


96  '  ALGEBRA.  [Eq.  SeC.  13. 

7.  A  and  B  owned  9800  acres  of  western  land.  A  sells 
J  of  his,  and  B  sells  ^  of  his ;  and  they  then  have  just  as 
much  as  each  other.     How  many  acres  had  each  1 

Ans.  A  4800  ;  B  5000. 

S,  A  son  asking  his  father  how  old  he  was,  received  the 
following  reply.  My  age,  says  the  father,  7  years  ago 
was  four  times  as  great  as  yours  at  that  time ;  but  7  years 
hence,  if  you  and  I  live,  my  age  will  be  only  double  of 
yours.    What  was  the  age  of  each  1 

Ans.  Father's  35  years;  Son's  14  years. 

9.  The  weight  of  the  head  of  Goliah's  spear  was  less  by 
one  pound  than  |  the  weight  of  his  coat  of  mail ;  and  both 
together  weighed  17  pounds  less  than  ten  times  the  spear's 
head.     What  was  the  weight  of  each  ? 

Ans.  Coat,  208  pounds;  Spear's  head,  25  pounds. 

10.  A  market  woman  bought  eggs,  some  at  the  rate  of 
2  for  a  cent,  and  some  at  the  rate  of  3  for  2  cents,  to  the 
amount  of  65  cents.  She  afterwards  sold  them  all  for  120 
cents,  thereby  gaining  half  a  cent  on  each  egg.  How  many 
of  each  kind  did  she  buy  ? 

Ans.  50  of  the  first  kind  ;  60  of  the  other  kind. 

11.  Says  A  to  B,  ^  of  the  difference  of  our  money  is 
equal  to  yours ;  and  if  you  give  me  $2,  I  shall  have  five 
times  as  much  as  you.     How  much  has  each  ? 

Ans.  A  $48  ;  B  $12. 

12.  A  and  B  possess  together  property  to  the  amount  of 
$5700.  If  A's  property  were  worth  three  times  as  much 
as  it  is,  and  B's  five  times  as  much  as  it  is,  then  they  both 
would  be  worth  $23,500.     What  is  the  worth  of  each  1 

Ans.  A  $2500;  B  $3200. 


THIRD  METHOD  OF  EXTERMINATION.  97 

13.  A  gentleman  has  two  silver  cups,  and  a  cover  adapt- 
ed to  each  which  is  worth  820.  If  the  cover  be  put  upon 
the  first  cup,  its  value  will  be  twice  that  of  the  second  ;  but 
i^  it  be  put  upon  the  second,  its  value  will  be  three  times 
that  of  the  first.     What  is  the  value  of  each  cup  ? 

Ans.  First  cup,  812  ;  second,  810. 

14.  Two  men  driving  their  sheep  to  market,  A  says  to 
B,  give  me  one  of  your  sheep  and  I  shall  have  as  many  as 
you.  B  says  to  A,  give  me  one  of  your  sheep,  and  I  shall 
have  twice  as  many  as  you.     How  many  had  each? 

Ans.  A,  5  sheep;  B,  7. 

0 

THIRD  METHOD  OF  EXTERMINATION. 

§129.  The  method  of  substitution  as  explained  in  the  last 
chapter,  may  be  modified  a  little.  We  will  show  how,  by 
using  question  1st,  in  the  last  section  of  equations. 

riM.  r  ^  ^+y  =100 

The  two  equations  were    ^  2x--dy=l50 
We  transposed  the  1st ;  thus,  07=100 — y 

Now,  before  we  substitute  the  value  of  x  for  x  itself  in 
the  second  equation,  we  will  transpose  the  second  equation 
so  as  to  make  x  stand  alone  ;  thus,  207=1 50 +3y. 

Then  substitute  the  value  of  x  as  found  before  by  the  first 
equation,  200 — 2y—l50-\-Sy  with  which  we  may 
proceed  as  before. 

§130.  Before  we  make  the  substitution  after  transposing, 
it  is  generally  best  to  find  the  value  of  x  alone  in  the  se- 
cond equation.     Thus, 

Given    Is^lgplO^  ^^  find  x  and  y. 


dS  AtGEBRAr  ■'         [Eq.  Sec.  14. 

JiU^'     ^;         Vlr^  23— 3V 

Transposing  and  dividing  the  1st,     x=  — ~- 

Transposing  and  dividing  the  2d,     x— -— ^. 

o 

Now,  as  it  is  evident  that  things  which  are  equal  to  the 

same^  are  equal  to  one  another  ;  one  value  of  x  is  equal 

to  the  other  value  of  x  ;  thus, 

23—32/  _10+22/ 

2  5~~ 

Destroying  the  fractions,  115 — 15i/=20+4y 

Transposing,  uniting,  and  dividing,  y=^ 

By  substituting  the  value  of  y  in  one  of  the  equations^ 

we  find  a?=4.     Whence  we  derive 

Rule  III.  to  exterminate  an  unknown  quantity. 

'§131.  Find  by  each  cfthe  equations,  the  value  of  that 
unknown  quantity  which  is  the  least  involved  ;  and  then 
form  a  new  equation  by  making  one  of  these  values  equal 
to  the  other, 

equations. — section  14. 

1.  Divide  $60  between  A  and  B,  so  that  the  difference 
between  A's  share  and  31,  may  be  to  the  difference  between 
31  and  B's  share,  as  6  to  7. 

Let    x=  A's  share  ;  and  i/=  B's. 

1.  By  the  first  condition,  x-\-y=60 

2.  By  tlje  second,  x — 31  :  31— 2/ :  •*  6  : 7 

3.  Multiplying  extremes  and  means,     7x — 217=186 — 6y 

4.  Transposing  the  1st,    .  a?=60 — y 

5.  Transposing  and  uniting  the  3d,  7a?=403 — 6y 

6.  Multiplying  the  4th,  7iP=420 — 7y 

7.  Making  5th  and  6th  equal,  403—6^=420—72^ 

8.  Transposing  and  uniting,  y=17 

9.  Substituting  17  in  the  4th,  ar=60— 17=43 

Ans.  A's  share  $43 ;  B's  $17. 


THIRD  METHOD  OF  EXTERMINATION.  ^9 

2.  There  is  a  fraction  such  that  if  1  is  added  to  the  nu- 
merator, its  value  will  be  \ ;  but  if  1  be  added  to  the  de- 
nominator, its  value  will  be  i.     What  is  that  fraction  1 
Let  ar=  numerator  ;  and  y  =  denominator. 

X 


1.  The  fraction  will  be. 

y 

2.  By  the  first  condition, 

x-\-l       1 

y       3 

3.  By  the  second, 

X           1 

2/  +  1      4 

4.  Multiplying  the  2d  hyy,  and  by  3, 

3:r  +  3=2/ 

5.  Multiplying  the  3d  by  y-^l,  and  by 

4,  ^x=y+l 

6.  Transposing  and  dividing  the  5th, 

4a:— 1=2/ 

7.  Making  4th  and  6th  equal. 

3a;+3=4iC— 1 

8.  Transposing  and  uniting. 

— a:=— 4 

9.  Substituting  the  value  of  3x  in  the  4th,       15=y. 

4 
Ans.  -. 

3.  What  two  numbers  are  those,  whose  difference  is  4 
and  5  times  the  greater  is  to  6  times  the  less,  as  5  to  4? 

Ans.  8  and  12. 

4.  There  is  a  certain  number,  consisting  of  two  places 
of  figures,  which  is  equal  to  4  times  the  sum  of  its  digits  ; 
and  if  18  be  added  to  it,  the  digits  will  be  inverted.  What 
is  tbat  number  1 

Let  iP=  first  digit  or  tens  ;  and  y=  the  units. 
10x-\-y=  the  number. 
^x-\-^y—  four  times  the  sum  of  digits. 
10a;+2/+18,=  when  18  is  added. 

lOy-\-x,=:  when  the  digits  are  inverted. 
By  the  first  condition,  10a?+2/=4a;+4i/ 

By  the  second,  10a?-f  i/-|-18=10«/+a; 

Ans.  24. 


100  ALGEBRA.  [Eq.  SeC.  14. 

5.  There  is  a  certain  number  consisting  of  two  figures  ; 
and  if  2  be  added  to  the  sum  of  its  digits,  the  amount  will 
be  three  times  the  first  digit;  and  if  18  be  added  to  the 
number,  the  digits  will  be  inverted.     What  is  the  number  ? 

Ans.  46. 

6.  A  person  has  two  snuflT-boxes  and  $8.  If  he  puts  the 
8  dollars  into  the  first,  then  it  is  half  as  valuable  as  the 
other.  But  if  he  puts  the  8  dollars  into  the  second,  then 
the  second  is  worth  three  times  as  much  as  the  first.  What 
is  the  value  of  each  ?  Ans.  First  $24  ;  second  $64. 

7.  A  gentleman  has  two  horses  and  a  chaise.  The  first 
horse  is  worth  $180.  If  the  first  horse  be  harnessed  to  the 
chaise,  they  will  together  be  worth  twice  as  much  as  the 
second  horse ;  but  if  the  second  horse  be  harnessed,  the 
horse  and  chaise  will  be  worth  twice  and  one  half  the  value 
of  the  first.  What  is  the  value  of  the  second  horse,  and 
of  the  chaise?  Ans.  Horse  $210  ;  Chaise  $240. 

8.  There  is  a  certain  number  consisting  of  two  digits. 
The  sum  of  these  digits  is  5  ;  and  if  9  be  added  to  the 
number  itself,  the  digits  will  be  inverted.  What  is  the 
number?  Ans.  23. 

9.  There  are  two  numbers  such  that  ^  of  the  greater 
added  to  4  of  the  less,  will  equal  13  ;  and  if  ^  of  the  less 
be  takep  from  ^  of  the  greater,  the  remainder  is  nothing. 
What  are  the  numbers  ?  Ans.  18  and  12. 

10.  There  is  a  number  consisting  of  two  figures.  If  the 
number  be  divided  by  the  sum  of  the  figures,  the  quotient 
will  be  4 ;  but  if  the  number  made  by  inverting  the  figures, 
be  divided  by  1  more  than  their  sum,  the  quotient  will  be  6. 
What  is  the  number  ?  Ans.  24. 

11.  There  are  two  numbers  such  that  the  less  is  to  the 


THIRD  METHOD  OF  EXTERMINATION.  lOl 

greater  as  2  to  5  ;  and  the  product  made  by  multiplying  the 
two  numbers  together  is  equal  to  ten  times  their  sum.  What 
are  the  numbers? 

Let  x=i  the  less ;  and  2/=»  the  greater. 

2v 

1.  By  the  first  condition,  ^—~R 

o 

2.  By  the  second,  xy=10x-\-10y 

Note. — If  we  wish  to  multiply  y  by  4,  we  put  4  imme- 
diately before  the  f/  as  a  co-efficient ;  and  in  the  same  way» 
if  we  multiply  y  by  x,  we  make  x  the  co-efficient  of  y. 

3.  Destroying  the  fraction  of  the  1st,     5x=:2y 

4.  Multiplying  by  2,  10x=4:y 

5.  Substituting  4y  for  10a?  in  2d,  xy=^y-\-10y 

6.  Dividing  by  y,  a:=4+10  =  14. 

Note. — When  we  divide  4y  by  4,  we  do  it  by  taking 
away  4;  when  we  divide  lOy  by  10,  we  do  it  by  taking 
away  the  10.  In  the  same  manner,  we  divide  by  y,  in 
taking  away  the  y. 

7.  Substituting  14  for  or  in  the  3d,        70 ^2y  .-.  2/=35. 

Ans.  14  and  35. 

12.  There  are  two  numbers,  whose  sum  is  the  ^  part  of 

their  product ;  and  the  greater  is  to  the  less  as  3  to  2.    What 

are  those  numbers?  Ans.  15  and  lO* 


102 


V     ^AI'GBBJIA. 


EQUATIONS  WITH  SEVERAL  UNKNOWN  QUANTITIES. 

§132.  When  there  are  three  or  more  unknown  quantities, 
they  are  exterminated  one  after  another  by  the  same  me- 
thods which  are  used  for  two  unknown  quantities. 

§1 33.  But  it  must  always  be  the  case  that  there  are  as 
many  independent  equations  as  there  are  unknown  quanti- 
ties.    See  §117. 

EXAMPLES. 

1.  Given  the  equations  <a^+22/+3z=16  \  ^^  ^naa;,y,^ 
(  x—y—2z=  —3  S  "" 

4.  Subtracting  3d  from  1st,  'Zy-\-^z—\'2 

5.  Subtracting  1st  from  2d,  '         y-(-2z=7 

6.  Multiplying  5th  by  2,  2y  +  43=14 

7.  Subtracting  4th  from  6th,  2; =2 

8.  Substituting  val.  of  z  in  5th,  i/=3 

9.  Substituting  in  the  1st,  a:=4. 


2.  Given  \  x-f  22/+3z=62  V  to  find  x,  y,  and  z. 


'0- 


2  '    3   '  4 

4.  Destroying  fractions  in  3d,  6a; +4t/+ 32=120 

6.  Subtracting  1st  from  2d,  y-\-2z=^S 

6.  Multiplying  1st  by  6,  6a?+6i/  +  6z=174 

7.  Subtracting  4th  from  6th,  22/+3z=54 

8.  Multiplying  5th  by  2,  2y+4z=66 

9.  Subtracting  7th  from  8th,  z=12 
Whence  by  substitution,  y=9  ;  and  a;=8. 


UNKNOWN  QUANTITIES.  103 

3.  Given   a7+3/+z=7  ;    2x — y — 3z=3;    and  ^^x — 3y 
+52=19  ;  to  find  x,  y,  z.  Ans.  a?=4, 2/=2,  z=l. 

4.  Given  x — y — 2=5;  3a?+4y+5z=52  ;  and  5a: — 4t/ 
— 32=32  ;  to  find  ar,  y,  and  2.      Ans.  a?=10, 2/=3,  2=2. 

5.  Given   Tar +5i/-f  22 =79  ;    8a?+7y+92=122  ;    and 
a?+4y+5z=55 ;  to  find  the  values  of  ^,  y,  and  2. 

Ans.  a?=4,  y=9,  2=3. 

6.  Given  a;+2/+2=13  ;  x+y-\-u=\l  \  x-\-z-\-u—lQ; 
and  3/+z  +  w=21 ;  to  find  the  values  of  x,  y?  z,  and  u. 

Ans.  a?=2,  3^=5,  2=6,  w=10. 


PART   II. 


LITERAL  ALGEBRA. 

GENERAL  PRINCIPLES. 

§134.  The  Algebraical  operations  which  we  have  hitherto 
treated  of,  belong  to  that  part  of  the  science  which  was 
known  to  the  ancients,  and  which  was  in  use  till  about 
A.  D.  1600.  About  that  time,  Franciscus  Vieta,  a  French- 
man, introduced  the  general  use  of  letters  into  Algebra, 
(denoting  the  known  quantities  in  a  problem  by  consonants, 
and  the  unknown  ones  by  vowels.) 

§135.  This  improvement  gave  a  new  aspect  to  the  science. 
So  that  now  the  object  of  it  is,  to  afford  means  for  disco- 
vering general  rules  for  the  resolution  of  all  questions  that 
can  be  proposed  relative  to  quantities. 

§136.  Operations  with  numbers  cannot  furnish  general 
rules ;  because  they  show  only  the  results  without  denoting 
how  they  were  obtained.  Thus,  12  may  be  the  result, 
either  of  multiplying  3  by  4,  or  adding  5  to  7,  or  of  sub- 
tracting 8  from  20,  or  of  dividing  48  by  4,  &c.  &c.  In 
order  that  a  result  should  furnish  a  rule,  it  should  not  de- 
pend upon  the  particular  values  of  the  quantities  which  we 
operate  with,  but  rather  upon  the  nature  of  the  question  ; 
and  should  also  be  the  same  for  every  question  of  the  same 
kind. 


GENERAL  PRINCIPLES.  105 

§137.  Algebra  enables  us  to  discover  rules,  because  it  is 
rather  the  representation  of  arithnnetical  results,  than  the 
results  themselves. 

§138.  In  the  first  place,  it  represents  the  quantities  under 
consideration  by  general  signs ;  that  is,  the  letters  of  the 
alphabet ;  which  may  stand  for  whatever  number  we  choose, 
and  even  for  as  many  different  numbers  as  we  choose,  one 
after  another. 

§139.  In  the  next  place,  by  different  methods  of  combin- 
ing these  representatives  of  numbers,  it  represents  the  ope- 
rations which  are  made  with  them  ;  and  this  is  done  in  such 
a  manner  that  the  quantities  still  retain  their  identity. 

§160.  And  further,  that  these  operations  may  be  facili- 
tated and  rendered  perspicuous,  many  of  them  are  repre- 
sented by  general  characters  called  signs.  The  following 
are  the  most  common. 

The  sign  +  (plus)  represents  addition. 

The  sign  —  (minus)  represents  subtraction. 

The  sign  x  (multiplied  by)  represents  multiplication. 

The  sign  .  is  sometimes  put  between  two  literal  quanti- 
ties instead  of  X  ;  as  a.6.  But  more  generally,  in  Algebra 
the  letters  are  joined  together,  to  represent  their  product. 

The  sign  -;-  (divided  by)  represents  the  division  of  the 
quantity  before  it  by  the  quantity  which  follows  it. 

Division  is  more  generally  denoted  in  Algebra  by  writing 
the  divisor  under  the  dividend  in  the  form  of  a  fraction. 

The  sign  =  (equals)  denotes  that  the  whole  quantity  on 
the  left  of  it,  is  equal  to  the  whole  quantity  on  the  right. 

The  sign  dr  or  =F  (plus  or  minus  and  minus  or  plus,) 
shows  that  either  by  addition  or  subtraction,  the  effect  will 
be  the  same. 


106  ALGEBRA. 

.  The  sign  ^  or  c/^  denotes  the  difference  between  two 
quantities. 

The  sign  >  or  <  {greater  than  or  less  than)  de- 
notes that  that  quantity  towards  which  it  opens  is  greater 
than  the  other. 

The  sign (a  vinculum,)  denotes  that  all  which  is 

put  under  it,  is  to  be  used  as  one  term. 

The  (  )  parenthesis  is  frequently  used  instead  of  the  vin- 
culum. 

The  sign  c»  {infinity)  denotes  a  quantity  that  is  infinitely 
large,  or  a  quantity  so  great  that  it  may  be  considered 
larger  than  any  supposable  quantity. 

The  cipher  0  is  sometimes  used  to  represent  a  quantity 
that  is  less  than  any  quantity  that  may  be  mentioned.  This 
's  always  the  case  when  the  cipher  is  used  as  a  denominator 
of  a  fraction. 

The  radical  sign  V  denotes  the  root  of  the  following 
quantity.  When  unaccompanied  by  a  figure,  it  represents 
the  square  root.  But  when  a  figure  is  put  over  it,  that  figure 
expresses  the  root  that  is  designed. 

A  co-efficient  \s  a  number  put  immediately  before  a  letter," 
as,  26. 

In  such  a  case,  the  co-efficient  is  called  a  numeral  co- 
efficient. Sometimes  when  one  letter  has  been  multiplied 
into  another,  the  first  written  one  is  called  a  literal  co-effi- 
cient ;  as,  ab. 

An  exponent  or  index  is  a  small  figure  placed  a  little  over 
and  a  little  to  the  right  of  a  quantity ;  as,  a^. 

An  exponent  may  be  either  positive,  negative,  or  frac- 
tional, as,  a^,  a~^,  a*. 

The  sign  :  :  represents  proportion,  and  :  denotes  the 
ratio  of  two  numbers. 

The  sign  oc  denotes  a  general  proportion. 


GENERAL  PRINCIPLES.  107 

A  simple  quantity  is  that  which  is  represented  by  one 
term. 

A  compovnd  quantity  is  that  which  is  represented  by  two 
or  more  terms. 

A  quantity  when  represented  by  one  term  is  sometimes 
called  a  nomial ;  with  two  terms  it  is  called  a  binomial ; 
with  three  it  is  called  a  trinomial ;  with  many  terms  it  is 
called  a  multinomial,  or  polynomial, 

§141.  The  algebraical  mode  of  designation  is  of  the 
greatest  use  in  universal  arithmetic ;  as  every  conclusion, 
and  indeed  every  step  by  which  it  is  obtained,  becomes  a 
general  rule  for  performing  every  possible  operation  of  the 
kind. 

Example  1.  Suppose  we  wish  to  divide  $660  between  a 
son  and  a  daughter,  so  that  the  daughter  shall  have  twice 
as  much  as  the  son.     What  must  we  give  to  each  ? 

Now,  a  question  similar  to  this,  and  with  the  same  numbers 
was  solved  in  the  First  section  of  Equations,  on  page  26. 
We  will  solve  this  in  the  same  manner,  with  the  exception 
of  using  a  instead  of  660. 

Stating  the  question,  x=.  What  the  son  has. 

2x=  What  the  daughter  has* 

Both  together  have  x-\-2x  ;  also  they  have  a  dollars. 

Forming  the  equation,  x-\-2x=a 

Uniting  terms,  3a?=a 

_,.-.-_  a 

Dividmg  by  3,  x=  — 

Here  we  find  that  the  son's  share  is  |  of  a,  which  at  this 
lime,  stands  for  $660. 

But  it  is  very  plain  that  we  would  solve  the  question  in  the 
very  same  manner,  if  the  sum  were  $240.  And  in  that 
case  a  would  stand  for  240  ;  and  the  son's  share  of  it  would 
be  $80,  and  the  daughter's  share,  $160. 


108  ALGEBRA. 

In  the  same  manner,  if  the  sum  were  ^360 ;  'then  a 
would  stand  for  $360,  and  the  son's  share  would  be  ^  of 
360  ;  that  is,  $120.  And  in  the  same  manner  we  may 
make  a  represent  any  sum ;  and  still  the  son's  share  of  it 
would  be  -J-  of  it.  Hence,  this  substitution  of  a  letter  for  a 
number,  is  called  generalizing  the  operation. 

We  see  that  this  result  has  given  us  a  general  rule  for  di- 
viding any  sum  between  two,  so  that  one  of  ihem  shall  have 
twice  as  much  as  the  other.  The  rule  is,  The  least  share 
shall  be  one-third  of  the  sum  ;  and  the  greatest  share,  two- 
thirds  of  it. 

§142.  In  the  same  manner,  we  may  generalize  all  the 
questions  in  the  First  Section  of  Equations.  That  is,  with 
each  question  we  may  find  a  rule  by  which  any  other  ques- 
tion like  it,  may  be  answered  with  fewer  figures  than  the 
Algebraic  operation  required.  But,  as  such  sums  are  not  apt 
to  occur,  there  will  be  no  practical  use  of  finding  rules  for 
them.  Notwithstanding,  as  the  exercise  may  be  instructive, 
the  teacher  may  require  his  pupils  to  go  through  them  if  he 
sees  fit. 

Example  2.  What  number  is  that,  which,  with  5  added 
to  it,  will  be  equal  to  40  ? 

This  is  the  first  problem  in  section  2,  which  we  will  gene- 
ralize ;  using  a  for  40,  and  b  for  5. 

Stating  the  question,  a;=  the  number. 

x-{-b=  after  adding. 
Forming  the  equation,  x-}-b—a 

Transposing  6,  x=a — b 

We  see  that  the  answer  is  found  by  subtracting  the  5 
from  the  40.     Thus,  40—5=35. 

The  third  question  is  similar  to  it ;  and  in  that,  a  repre- 
sents 23,  and  b  represents  9.    The  answer  is  23 — 9=14. 


GENERAL  PRINCIPLES.  109 

Example  3.  Divide  $17  between  two  persons,  so  that 
one  nnay  have  $4  more  than  the  other.     [Prob.  4.  Sect.  2.] 

Represent  17  by  a,  and  4  by  b. 
Forming  the  equation,  x+x-^-b^a 

Transposing  b,  x-\-x=a — b 

Uniting  terms,  2x=a — b 

Dividing  by  2,  x=a — b 

~2~ 

The  answer  is  found  by  subtracting  the  difference  or  4, 
from  the  whole  sum,  and  then  dividing  by  2. 

And  this  is  the  rule  for  all  similar  sums. 

The  5lh  question  is  similar  to  it :  and  in  that,  a  represents 
55,  and  b  represents  the  difference  or  7.  It  is  found  by 
the  rule  just  shown. 

Thus,  523!  =  !?=, 4. 

-i  lit 

Perform  the  7th  and  lOih  by  the  same  rule. 

§143.  As  this  rule  is  of  some  importance,  it  will  be  well 
to  remember  it.  If ^  from  a  number  to  be  divided  into  two 
parts  J  we  subtract  the  difference  of  those  parts,  half  the 
remainder  will  be  equal  to  the  smaller  part. 

Example  4.  In  the  same  questions,  let  us  take  x  for  the 
greatest  share.     Then  x — b  =  the  less. 
Forming  the  equation,  x-\-x — &=a 

Transposing  and  uniting,  2a*==a+6 

Dividing  by  2,  x=a+5 

§144.  Here  we  have  another  rule.  If,  to  a  number  to 
be  divided  into  two  parts,  we  add  the  difference  between 
those  parts,  half  the  sum  will  be  equal  to  the  greater  part, 

10 


110  ALGEBRA. 

Find  the  greater  part  in  questions  4,  5, 7,  and  10,  by  this 
rule. 

§145.  The  mere  letters  in  the  answer  of  an  algebraical 
operation  are  called  di  formula.  They  are  not  called  a 
rule^  until  they  are  turned  into  common  language. 

Example  5.  The  learner  may  now  generalize  question  6; 
and  by  the  formula  that  he  obtains,  he  may  find  the  an- 
swers in  questions  8,  9,  and  11.  The  two  differences  will 
be  h  and  c. 

We  have  said  that  in  Algebra  the  arithmetical  operations 
on  nuqibers  are  only  represented  by  different  methods  of 
combining  the  signs  that  stand  for  those  quantities.  And 
now,  although  we  have  shown  in  our  progress  thus  far,  what 
some  of  those  methods  are,  it  may  be  well  to  review  them  a 
little. 


ADDITION  AND   SUBTRACTION  OF  ALGEBRAICAL 
QUANTITIES. 

§146.  One  algebraical  quantity  is  added  to  another  by 
writing  one  quantity  after  the  other,  taking  care  to  put  the 
sign  plus  +  between  them.  Thus,  a+/ — c  is  added  to 
d — c+^j  so  as  to  make  d — e-\-h-\-a-\-f — c.  Or,  as  it  is 
easier  to  read  the  letters  in  their  alphabetical  order,  their 
sum  may  be  written  a-\-b — c-\-d — e-f/» 

§147.  One  algebraical  quantity  is  subtracted  from  an- 
other, by  changing  the  sign  or  signs  of  the  quantity  which 
is  to  be  subtracted,  and  then  writing  that  quantity  after  the 
other.    Thus,  a+A — y  is  subtracted  from  b — x-fc,  by  first 


ALGEBRAICAL  QUANTITIES.  Ill 

making  it  — a — A+y,  and  then  writing  the  whole  quan- 
tity,  b — x-\-c — a — h-\-y  \  or,  h — a-f-c — h-\-y — x. 

§148.  When  quantities  that  have  been  nnade  by  addition 
or  subtraction,  have  like  terms  in  them,  they  may  be  re- 
duced to  smaller  expressions^  by  uniting  the  like  terms. 
This  is  done  by  putting  into  one  part,  all  those  which  have 
the  sign  +,  and  into  another  part,  all  those  which  have 
the  sign  —  ;  then  subtract  the  least  result  from  the  greatest, 
and  give  to  the  remainder  the  sign  that  belonged  to  the 
greatest  result. 

§149.  In  uniting  the  terms  of  compound  numbers,  we 
consider  the  literal  part  of  the  term  as  a  unit ;  thus,  2a 
and  3a,  are  regarded  as  2  units  and  3  units  of  a  particular 
kind,  which  when  put  together,  make  5  units  of  that 
kind.  Now,  we  have  seen,  §139,  that  the  co-efficient  of  a 
quantity  may  also  be  literal ;  as  in  6a,  ca,  &c.  In  such 
cases,  the  whole  term  ha  or  ca  becomes  the  unit,  each  of  a 
different  kind ;  and  of  course  are  not  like  quantities  and 
cannot  be  united. 

§150.  But  if  there  are  several  similar  units  of  this  kind, 
they  may  be  united  by  the  general  rule.  Thus,  ha — ca 
-^ha-\-ca-\-ha-\-ca,  can  be  united  into,  Zha-\-ca.  ax — hx 
-{■ax-\-2bx — t^ax-\-hx,  are  equal  to,  — ax-\-2hx  \  ov  2hx 
— ax, 

§151.  Again,  we  have  seen,  §65,  that  several  quantities 
are  sometimes  united  by  a  vinculum.  In  such  cases  all  that 
is  embraced  by  the  vinculum,  is  regarded  as  a  unit  of  that 
kind ;  and  may  have  a  co-efficient.  Thus,  in  the  expres- 
sions, 3xa — h-\rx^  and  b(x-\-ax — i/),  a — h-\-x  is  a 
quantity  taken  3  times,  and  x-\-ax — y  is  a  quantity  taken  5 
times.  Like  quantities  of  this  kind  can  be  united  ;  thus, 
2{ay — hx-\-x)-\-b{ay — hx-]-x)  =  l{ay — hx-\-x). 


112  ALGEDRA. 

§152.  In  uniting  terms,  great  care  must  be  taken  that 
the  literal  part  be  entirely  alike.  Thus,  2bx-\-ZcXj  cannot 
be  united.  Neither  can  Sy — 2ay ;  nor,  6(a-\-bx)-\-2{ax 
■i-bx) ;  nor,  3.  ay — by  +  2.ay^ — by,  nor,  A{ax — bx) — 
2(ax-\-bx)  ;  neither  in  any  other  case  where  there  is  the 
least  difference  in  any  part  but  the  leading  co-efficient. 

EXAMPLES. 

Unite  the  following  quantities. 

1.  Sax  —  2y-\-4:X  —  oy-\-ax  —  Sy,     Ans.  8aar — lOy 

2.  Sx-^ay  —  2x  —  ay-i-^x-\-Say  —  2x+Aay. 

Ans.  ^x-^-lay 

3.  Aax  —  y+Say — 2  —  2ax-\-ay  —  ly-\-Q-\-2ay-\-y. 

Ans.  2ax-\-^ay  —  7^-4-6 

4.  ax  —  ay^ —  Say-\-bax  —  2ay-\-lay  —  Aax —  Say^ 

Ans.   2ax-\-2ay  —  9«y^ 

6.  3(a  -  y)  +  4(a  —  y)  +  2(«  —  y)  +7(a  —  y). 

Ans.  16(a  —  y) 

6.  _4(a+&)+3(a+6)— 2(a  +  6)+7(a-f&.) 


Ans.  4.  a-\-b 

7.  2{ab-\'X)-it-S{ax+b)  —  A{x^y)  —  2{ab-\-x) 

Ans.  S{ax-\-b)  —  4(a:  — y) 

8.  7y  —  4(fl  +  ft)  +  62/  +  22/  +  2(a  +  6)  +  (a  +  6)  -f  y 
—  3(a  +  b.)  Ans.  16y  —  4(a+6) 

9.  x^  -\-ax  —  ab-\-ab  —  x^-^xy-\-ax-\-xy — A.ab-\-x^-\- 
x^  —  x-^xy-\-xy+ax.    Ans.  2x^-\-Sax — Aab-\-\xy  —  x 

10.  X+12-— aa;+2/  — (48  — a:  — ax+3.y). 

Ans.  2x  — 36  — 2i/, 


MULTIPLICATION  OF  ALGJ^RAICAL  QUANTITIES.      113 

11.  ab  —  Axy  —  a~x^  —  (2xy  —  b-\-lAx-\-x^) 

Ans.  ah  —  6xy — a  +  b  —  140^ — 2x^^ 


12.  S{x-\-y)-\-{^.x-^y).  Ans.  l{x-\-7j). 

VS.  2{a--b)  —  x—{'S.a+b  —  x'').  Ans.  x''—{a+b)—x. 


14.  From  4.a+6,  take  a+b  —  S.x  —  y. 

Ans.  S.a-\-b-{-S.x  —  y. 

15.  a-^b  —  {2a  —  .36)  —  {5a-\-7b)  —  ( —  lSa+2b). 

Ans.  7a —  56. 

16.  37a  —  ox  —  (3a  —  26  —  5c)  —  (6a  —  46  +  3A). 

Ans.  28a+66  — 5a:  +  5c— .3/t. 

EXERCISES  IN  ECiUATIONS. 

§153.  The  learner  may  now  generalize  the  problems  in 
Sections  3  and  4  of  Equations,  pages  35,  and  40.  This  he 
can  easily  do,  if  he  takes  care  to  make  one  of  the  first  let- 
ters of  the  alphabet  stand  for  each  numeral  quantity.  When 
the  same  numeral  quantity  occurs  more  than  once  in  the 
same  question,  the  same  letter  must  stand  for  it  each  time. 


MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES. 

§154.  We  have  shown,  §29,  that  a  literal  quantity  may 
be  multiplied  by  writing  the  multiplier  before  that  quantity. 
This  is  the  case  whether  the  multiplier  is  numeral  or  literal. 
Thus,  a  limes  x,  is  written  ax;  bxc=bc.  In  the  same 
manner,  a  times  be  becomes  a6c ;  andy  times  a6c  becomes 
fabc.  As  fxabc  is  the  same  as  abcxf,  we  see  that  it  is 
of  no  consequence  what  order  we  make  of  the  letters  in  th« 
10* 


114  ALGEBRA. 

product,     ahcd  =  acdb  =  cadby  &c.     But  it  is  generally 
more  convenient  to  follow  the  order  of  the  alphabet. 

§155.  Therefore,  to  multiply  one  simple  quantity  by  an- 
other, write  the  quantities  one  after  another,  without  any  sign 
between  them.  Thus,  abx  times  c/v/  =  abxcfy;  bax  times 
cdf=^  bacdfx.  But,  if  there  are  more  than  one  numeral 
co-efficientj  those  co-efficients  must  be  multiplied  as  in 
arithmetic,  and  placed  before  the  product  of  the  literal 
quantities.  Thus,  3a  x  2a?  =  Z.2ax  =  Qax,  2bc  x5rs  = 
lObcrs. 

EXAMPLES. 

-  1.  Multiply  a  by  b.  Ans.  ab, 

2.  Multiply  ab  by  c.  Ans.  abc, 

3.  Multiply  ab  by  cd.  Ans.  abed. 

4.  Multiply  2acx  by  by.  Ans.  2abcyx, 

5.  Multiply  "ibrs  by  2mnx.  Ans.  Qbmnrsx. 

6.  Multiply  2adn  by  5cmx, 

7.  Multiply  ^rsyop  by  4an/x. 

8.  Multiply  2abcx  by  8abrx, 

§156.  By  the  foregoing  principle,  axa  —  aa.  Now, 
as  in  Algebra  the  same  factor  is  often  found  two  or  more 
times,  Stifelius  adopted  a  method  for  shortening  such  ex- 
pressions,* in  which  he  has  ever  since  been  followed.  The 
method  is  this  ;  when  the  same  letter  enters  as  a  factor  two 
or  more  times  into  any  quantity ,  we  write  the  factor  but 
oncej  and  put  at  the  right  of  it  and  a  little  raised,  a  figure 
denoting  how  many  times  it  has  been  multiplied.  Thus, 
aa  is  written  a^ ;  bbb  is  written  6^ ;  xxxx  is  written  ar* ; 
aabbbyyyy  is  written  a*  b^  y*. 

•  A,  D.  1554. 


i^^i 


MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.      115 

§157.  Mathematicians  are  accustomed  to  call  aa  or  a", 
the  second  power  of  a,  or,  a-second  'power ;  a*,  is  called 
a-third  power,  &c. 

§158.  The  figure  that  denotes  the  power  of  any  quantity 
is  called  the  exponent  or  index  of  that  quantity. 

§159.  All  quantities  are  said  to  have  an  exponent,  either 
expressed  or  understood.  Thus,  a  is  the  same  as  a* ;  6=6* ; 
&c.  The  written  exponent  affects  no  letter  except  the  one 
over  which  it  is  written ;  unless  it  is  denoted  by  a  vincu- 
lum. 

§160.  Great  care  must  be  taken  by  the  pupil  not  to  con- 
found the  co-efftcient  with  the  exponent  ;  as  their  effects  are 
entirely  different.  The  co-efficient  shows  addition,  the  ex- 
ponent denotes  multiplication.  For  example,  if  a=5,  then 
3a=54-5-f5=15;  but  ^3=5x5x5  =  125. 

§161.  We  have  seen  X\\aX  a^=^aa  ;  and  that  a^^^aaa; 
now  aaxaaa=aaaaa.  So  we  see  that  a^xa^=a^.  Hence 
we  establish  the  rule  that  when  both  multiplier  and  multi' 
plicand  are  denoted  by  the  same  letter,  their  product  is 
found  by  adding  their  exponents.  X^Xx*=x^,  y^Xy^=s 
y' ;  &c. 

EXAMPLES. 

9.  Multiply  2amn  by  a.  Ans.  2a^mn. 

10.  Multiply  Sabcx  by  4ax.  Ans.  12a^bcx^, 

11.  Multiply  5bcmn  by  'Sbc,  Ans.  15b^c^mn, 

12.  Multiply  6a'^xy  by  ^ax^y.  Ans.  2Aa^x^y^ 

13.  Multiply  Aa^bcx  by  a^b^c,  Ans.  Aa^b*c^x. 

14.  Multiply  'Sa^m^  by  a'^m^n. 

15.  Multiply  ba^x^  by  4a^a?*. 

16.  Multiply  2771  Wic  by  9aTa?*rn». 


116  ALGEBRA. 

17.  Multiply  Sb^c^'n^  by  SaH^mn^. 

18.  Multiply  7aya;*  by  \2a'>¥x. 

19.  Multiply  ^ni'x^a'  by  la^'h^m, 

20.  Multiply  \2aJ^mHx^  by  ISb^cn^y. 

§162.  When  one  of  the  factors  is  a  compound  quantity, 
(§61  and  62,)  we  multiply  each  term  by  the  other  factor, 
and  set  down,  their  products,  each  with  their  proper  sign. 
In  doing  this,  we  generally  begin  at  the  left ;  thus, 

a-{-bc  ab — c. 

Multiplied  by  x  Muhiplied  by  a 

ax-\-bcx  a% — ac.  , 

21.  Multiply  ax-\-bx  by  xy.  Ans.  ax^y-\-bx!^. 

22.  Multiply  2d^bc  —  rxy^  by  "lay.  Ans.  Aa^bcy — 2arxy*. 

23.  Multiply  7ay+l  by  Sr.  Ans.  2lary+3r. 

24.  Multiply  2xy+ab-\-c*  by  2ax. 

25.  Multiply  5  —  7x-{-2a^b  by  4acy 

26.  Multiply  2m"n —  Srp-{-2sx^  by  6a^ba. 

27.  Multiply  2en  —  Aan+5  by  120^^x5. 

28.  Multiply  Ay-\-y^  by  2xy. 

29.  Multiply  3a  —  46+4  by  6y, 

30.  Multiply  6ax''-^a^—  1  by  2ax\ 

31.  Multiply  4  + 3a  —  ic^  by  ay. 

32.  Multiply  2a+5i2_^3c  — 5e  by  3a». 

33.  Multiply  7b^  —  4-ha«a? — x"^  by  4a«a?. 

34.  Multiply  6a;  -f-7a  —  axy  —  2y  by  Zbx^. 

§163.  As(a+6)  y.x=^ax-\-bx\  weknow thatar  x(a-|-6) 
also  =ax+bx.  Whence  we  see  that  if  we  wish  to  multi- 
ply X  by  a-\-b ;  we  first  find  the  product  of  a  times  a?,  and 
then  add  to  it  the  product  of  h  limes  x.     And  generally^ 


MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.      117 

when  the  mvltiplier  is  composed  of  several  terms^  the  pro- 
duct  is  made  up  of  the  sum  of  the  products  of  the  multipli- 
cand hy  each  term  of  the  multiplier.    Thus,  {x-{-y)x{a-\-b) 

=  ^  a  >  and  <b  c^^  i  o+h 

f  ax-f-ay )  \bx-^byj       \  ax-\-ay-\-bx-\-bu, 

EXAMPLES. 

35.  Multiply  2a  +  36  by  4a+56. 
2a-\-'Sb 
4a +56 
8tt*+12a6 

4-10a&+156« 


8tt2+22a6  +  1562 

36.  Multiply  a+ft  by  a-^b.  Ans.  a2+2a6-f  6*. 

37.  Multiply  a-f  6~c  by  a'^b, 

Ans.  a^-f-a'^i  — aV+a6+62  — 6c. 

39.  Multiply  x"-\-2xy-{-'f  by  x-\-y. 

Ans.  a;'  +  3a:=i/-t-3a??/2+2/^ 

39.  Multiply  a  —  a?  by  2a  +  3a7.  Ans.  2a^-\-ax  —  Zx^. 

40.  Multiply  X  —  4  by  x-\-S.  Ans.  x''+^x  —  32. 

41.  Multiply  X  —  y  hy  x  +  y,  Ans.  x^  —  y*. 

42.  Multiply  Aa^  —  Sxy^+7ax  —  ay  by  2aHxY 

43.  Multiply  a^'bc^+Sa^c^  —  bb^'c  by  aa^^+^/x^. 

44.  Multiply  m^a^-^r'^n^b'' —l—m^n  by  2m86'»+4a«/i*. 

45.  Multiply  Sa'^x^y* — 2a*x^y^-\-a^\f  by  3a'»m+2wy*. 

§164.  As  (a — x)y^b  =ab  —  a-6;  we  also  know  that  6  X 
(a — x)=ab  —  xb.  Therefore  to  find  a  —  a:  limes  6,  we  first 
find  the  product  ofa  times  6,  and  then  subtract  from  it  x  times 
b.  This  principle  is  very  plain  in  figures.  Thus,  suppose 
we  have  7  times  8,  and  wished  only  4  times  8 ;  if  we  take 


i 


118  ALGEBRA. 

3  times  8  away  from  the  7  times  8,  we  should  obtain  4 
times  8  ;  as  56 — 24=32.  Whence  we  have  the  general 
rule,  if  there  is  a  negative  quantity  in  the  mnltiplier  ;  the 
•product  of  that  quantity  and  the  multiplicand  must  he  sub- 
tracted from  the  product  of  the  positive  quantities  and  the 
multiplicand.     Thus,  {a-\-h)  X  (c — d). 

Ca-\-b      ^  icL+b      ^        a-\-b 

5=  7  c  >  minus  <  <i  >  or    c — d 

\ac-\-bc  y  \ad-j-bdl        ac-\-bc — ad — bd. 

Here  we  see  that  the  product  of  fZ  into  a 4-^  is  subtracted 
from  the  product  of  c  intoa  +  6 ;  and  therefore  the  signs  of 
ad-\-bd  are  changed  to  — ad — bd. 

§165.  By  examining  the  answer  of  this  last  example,  we 
shall  observe  a  principle  which  will  enable  us  to  be  more 
rapid  in  the  multiplying  operation.  It  is  this;  when  we 
multiply  a  +  term  by  a  +  term\  the  product  in  the  answer 
is  a  +  term ;  and  when  we  multiply  a  +  term  by  a  — 
term^  the  product  in  the  final  answer  is  a  —  term.  It  will 
be  well  for  the  pupil  to  explain  this. 

By  understanding  this  principle,  we  are  able  to  set  the 
final  answer  down  at  first ;  as  we  have  only  time  to  remem- 
ber that, 

§166.  When  the  signs  are  alike,  the  product  is  +. 

§167.  When  the  signs  are  unlike,  the  product  is  — . 

EXAMPLES. 

46.  Multiply  a-\-b  by  a—b. 
a-\-b 
a—b 

—ab—b^ 


Ans.  a^- 


-4i%'. 


MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.      119 

47.  Multiply  x+8z  by  dx''—7xz. 

Ans.  3a:3+17»^z— -56a;2^ 

48.  Multiply  x^+xy  +  y^  by  x — y.     Ans.  x^ — y^. 

49.  Multiply  2x-\-3y  by  'Sx—^y,  Ans.  Gaj^-f  a:^ — 12^^ 
.50.  Multiply  b^-^b^x+bx^+x^  by  b — x.  Ans.  b* — x*, 
51.  Multiply  a — b  by  c — d, 

a — b     1  C  a — b      ^        a — b 

c  >  minus  <d  >  or    c — d 


ac — be  J 


ac — be  J  V  ad — bd  J        ac — be — ad-\-bd. 

Here  we  see  that  in  subtracting  d  times  a — b,  we  change 
the  signs  of  ad — bd  to  — ad-{-bd. 

§168.  Whence  we  learn, 

-f  multiplied  by  +,  produces  + 
+  multiplied  by  — ,  produces  — 

—  multiplied  by  +»  produces  — 

—  multiplied  by  — ,  produces  -f. 

And  by  remembering  this  we  can  always  set  down  the 
Jinal  answer  at  first. 

52.  Multiply  a — x  by  a — x, 

a — X 
a — X 


ix-\-x^ 


tt« — 2ax+x'^. 

53.  Multiply  2x  —  Sa  by  4a?  —  5a. 

Ans.  Sx^  —  22aa;-f  15a». 

54.  Multiply  2a —  5y  hy  a  —  2y.    Ans.  2a*  —  Qay-^-lOy^. 

55.  Multiply  a^+ac  —  c^  by  a  —  c.     Ans.  o*  —  2ac'^-\-c', 

56.  Multiply  a-\-b-^dhy  a —  b. 

Ans.  o"  —  ad  —  b^-j-bd. 


k 


120  ALGEBHA. 

57.  Multiply  Ax  — 5a  — 2b  by  Sx  —  2a-\-5b. 

Ans.  l2x^  —  2Sax-\-lAbx+\0a^  —  21ab^l0b^. 

58.  Multiply  x^  —  y^  —  z^  by  x — y  —  z, 

59.  Multiply  6+xy  —  a^  —  mif  by  a^  —  Sx^-^y*, 

60.  Multiply  2a^b  —  3ac2-f-46V  —  1 

by  2a^c^  -—  5b^c  —  8a*. 

§169.  In  order  to  facilitate  the  practice  of  multiplication, 
it  is  best  to  observe  the  following  method.  First,  determine 
the  sign,  then  the  co-efficient,  then  the  letters  in  their  order, 
and  then  the  exponents. 

Gdneral  Properties  of  Numbers. 

§170.  We  have  before  stated  that  algebraical  operations, 
§137,  by  reason  of  the  quantities  themselves  being  retained 
in  their  original  value,  do  show  us,  in  their  results,  import- 
ant general  principles.  We  will  here  make  a  few  naultipli- 
cations  of  some  quantities,  whose  results  show  us  some  re- 
markable general  properties  of  numbers.  These  properties 
the  pupil  should  remember,  as  they  are  of  frequent  use  in 
the  subsequent  parts  of  this  study. 

§171.  Suppose  we  have  two  numbers,  a  and  6,  of  which 
a  is  the  greatest.  Then  their  sum  =  a-\-b  ;  and  their  dif- 
ference =  a —  b.     Then  a-i-b 

a — b 
aFTab 
—ab—b^ 


a^—b^ 
By  this  operation,  we  find  the  general  property  of  num- 
bers which  it  would  be  difficult  to  find  by  any  arithmetical 
operation.  It  is  that,  if  we  multiply  the  sum  of  tico  num- 
bers by  their  difference,  the  product  will  be  the  difference  of 
the  squares  of  those  numbers. 


^n 


MULTIPLICATION  OF  ALGEBRAICAL  QUANTITIES.     121 


§172. 

Again, 

take  the 

same  quantities. 

and 

multiply 

their 

sum,  by 

their  sum* 

a+6 

a-\-b 

a^+ub 

+  ab-\-b^ 

a^+2ab-\-b^ 

By  this  operation  we  find  the  following  general  property. 
2'he  square  of  the  sum  of  two  numbers  is  equal  to  the 
square  of  the  first  number,  plus  twice  the  product  of  the 
two  numbers,  plus  the  square  of  the  last  number, 

§173.  Again,    take   the  same  quantities,  and   multiply 
their  difference,  by  their  difTerence. 
a — b 
a—b 
a^—ab 

—ab-\-b^' 
a^—2ab-{-b» 
Therefore,  the  square  of  the  difference  of  two  numbers, 
is  equal  to  the  square  of  the  first  number,  minus  twice  the 
product  of  the  two  numbers,  plus  the  square  of  the  second. 
§174.  The  only  difference  between  the  square  of  the  sum, 
and  the  square  of  the  difference,  is  in  the  second  term  ; 
being  in  one,  positive,  and  in  the  other,  negative.     Let  us 
subtract  one  from  the  other. 

a^-^2ab+b» 

a^—2ab+¥ 

Aab 

The  actual  difference  between  the  square  of  the  sum,  and 
the  square  of  the  difference,  is  four  times  the  product  of 
the  two  numbers. 

11 


122  ALGEBRA* 

^175.  If  when  the  sum  of  two  quantities  has  been  raised 
to  the  second  power  and  the  co-efficient  of  the  second  term 
has  been  rejected,  the  quantity  thus  obtained  be  multiplied 
by  the  difference  of  the  two  original  quantities ;  the  result 
will  be  the  difference  of  the  third  powers  of  the  two  quart' 
titles.  Also,  if  we  perform  the  same  operation  with  the 
difference  of  the  two  quantities,  multiplying  by  their  sum, 
we  shall  obtain  the  sum  of  the  third  powers  of  the  two 
quantities.     Thus, 

a+h  a — h 

a-\-h  a — b 


Rejecting  the  co-efficients, 

a^+ab  +  b^  a^—ab-j-b" 

a — b  a+b 

a^  +  a^'b+ab^  a^—a^b+ab^ 

^c^b—ab^—b^  +  a^b—ab^ + b^ 


§176.  It  is  often  the  case  that  it  is  better  to  denote  the 
multiplication  of  compound  quantities,  than  to  perform  it  j 
on  account  of  operations  that  follow.  Thus,  a — h  times 
ic+y,  may  be  written  (a — b).  (x-{-y).  No  general  rule 
can  be  given  to  determine  when  one  method  is  preferable 
to  the  other.  Experience  is  the  best  teacher  in  this  par- 
ticular.    But  it  was  thought  best  to  mention  it  in  this  place. 

When,  after  the  multiplication  has  been  denoted,  the  seve- 
ral terms  are  actually  multiplied,  the  expression  is  said  to  be 
expanded. 


aiVISION  OF  ALGEBRAIC  aUANTITIES.  128 


DIVISION  OF  ALGEBRAIC  QUANTITIES. 

§177.  Division  may  be  represented  by  the  sign  -r-  ,  as 
«  -H  6,  is  read  a,  divided  by  h  ;  (a+6)  H-  (c — d)y  is  read 
a-\-h,  divided  by  c — d.  But  the  most  usual  way  to  denote 
division  (§67,)  is  to  write  the  divisor  underneath  the  tiivi- 

dend  :  thus,  7-,        — :;. 
b  c — a 

§178.  But  it  often  happens  (§71,  73,)  that  the  fraction 

made  by  this  representation  is  an  improper  fraction ;  and 

one  in  which  the  numerator  can  be  actually  divided  by  the 

denominator.     In  such  cases,  it  is  generally  best  to  pe  rfom 

the  division.     We  have  always  done  so  in  the  former  part 

lOrc 
of  this  treatise.     Thus,  4a;  divided  by  2=2a: ;   — -  =  2x, 

o 

§179.  Let  us  first  look  at  the  case  where  the  same  quan- 
tity is  in  both  the  dividend  and  the  divisor.  7a-r-7=a  ; 
126-r-12=6.  In  the  same  manner,  ab-r-a^=b\  dc-7'd=^c^ 
This  may  be  easily  proved.  For,  ab  is  the  product  of  a 
into  b ;  and  of  course  if  we  divide  by  what  was  the  multi- 
plier, we  shall  obtain  the  old  multiplicand  again;  as  may  be 
seen  by  trying  the  product  of  any  two  numbers. 

§180.  Whence  we  derive  the  general  rule,  that  when  the 
divisor  is  found  as  a  factor  in  the  dividend,  the  divisionis 
performed  by  erasing  that  factor  from  the  dividend, 
<imn-T-a=mn,  because  amn^==a  times  mn,  amn-7-m=an, 
because  amn=m  times  an,  amn-i-n^^am,  because  amnis 
n  times  am. 


124 


ALGEBRA. 


EXAMPLES. 

1.  Divide  8c  by  8.  Ans.  c 

2.  Divide  be  by  b,  Ans.  c. 

3.  Divide  Im  by  7.  Ans.  m. 

4.  Divide  am  by  a.  Ans.  wj. 

5.  Divide  6rf  by  b.  ^ 

6.  Divide  7a6  by  7. 

7.  Divide  cab  by  c. 

8.  Divide  bad  by  6. 

9.  As  ahd  is  the  same  product  as  the  last,  divide  that 
by  6.  Ans.  ad. 

10.  Divide  cde  by  c. 

11.  Divide  the  same  product  in  another  form,  thus,  dec 
by  c.  Ans.  (/c. 

12.  Divide  abc  by  c. 

13.  Divide  abc  by  6. 

§181.  As  a^=:aa,  it  is  evident  that  if  we  divide  a^  by  a 
the  quotient  is  a ;  because  we  take  away  one  of  the  written 
a's.  So,  if  we  divide  a^  by  a,  the  quotient  is  a*  ;  because 
a'  is  the  same  as  aaaaa,  which  -;-  cf,  gives  aaaa  or  a*.  So, 
if  we  divide  a^  by  a^,  the  quotient  is  aaa  or  a^.  And  if  we 
divide  a^  by  a*,  the  quotient  is  a ;  because  aaaaa-r-aaaa 
=a.  Hence  we  see,  that  when  there  are  exponents  in  either 
the  dividend  or  divisor,  the  division  is  performed  by  sub- 
tracting the  exponent  of  the  divisor  from  the  exponent  of 
the  dividend.     b^-h-b^=^b^ ;  x''-^x^=ix*. 

§182.  As  every  literal  quantity  is  understood  to  have  the 
number  1  for  its  co-efficient,  it  is  evident  that  if  we  divide 
by  1,  the  quotient  would  be  the  same  literal  quantity. 
Thus,  a-j-l=a.  And  again,  if  we  divide  a  literal  quan- 
tity by  itself,  the  quotient  will  be  1.  Thus,  la-j-a=l  ; 
the  divisor  being  a  factor  in  the  dividend. 


DIVISION  OP  ALGEBRAIC  QUANTITIES.  125 

§183.  It  sometimes  happens  that  the  co-efficient  contains 
the  divisor  as  a  factor.  Thus,  8«  is  the  same  as  2  times 
4a,  or  4  times  2a ;  and  therefore  can  be  divided  by  2  or  by 
4.  In  the  same  manner  8ab  may  be  divided  by  2«,  or  4fif, 
or  2b,  or  46  ;  because  it  is  2a  times  46,  or  4a  times  26. 

We  have  only  to  remember  to  take  those  factors  out  of 
the  dividend  which  are  equal  to  the  divisor.  And  in  gene- 
ral, when  there  are  co-efficients  in  both  the  divisor  and  the 
dimdend,  divide  the  co-efficient  of  the  dividend  by  the  co- 
efficient of  the  divisor;  and  then  proceed  with  the  literal 
quantities  as  before  directed,  10a6c-i-56=2ac ;  I2a^xy 
-r-2a^y=4:ax, 

EXAMPLES. 

14.  Divide  a^  by  a°.  Ans.  a. 

15.  Divide  x^  by  a?".  Ans.  a?*. 

16.  Divide  a^b'^y  by  a^b.  Ans.  ab^y. 

17.  Divide  d^c^x'^  by  dc^x*.  Ans.  d^c^x^. 

18.  Divide  a*m^x  by  a*m^, 

19.  Divide  a*x^y^  by  ax*y^. 

20.  Divide  d^y''  by  dy , 

21.  J)W\AepH'^sH  by  r^sU 

22.  Divide  ab^C'd^  by  ab^cd\ 

23.  Divide  ax'^y^  by  ax^y^* 

24.  Divide  c^r^sHx^y^  by  c^r^ly^. 

25.  Divide  6a=^6c''  by  26c^  Ans.  Ba^c^ 

26.  Divide  12ax^y*  by  4a2/«.  Ans.  3a:^y^. 

27.  Divide  2\bc^xy^  by  3c2(^ 

28.  Divide  42c7(/«a;^  by  6c=^(/^a?. 

29.  Divide  36jor^s/*  by  4rs^^ 

11* 


m 


ALGEBRA. 


30.  Divide  ^^^m^n^x^y  by  9m^ny. 

31.  Divide  66a''c''dx^  by  3aVa;^ 

32.  Divide  ^Sc^'r'^x^y^  by  Scr^xyK 

33.  Divide  TZa^r^m"  by  ISa^rm^ 

§184.  We  have  shown,  §162,  that  (a+&)Xc=«c4-6c. 
Of  course  ( /c4-6c)-r-c=a4-^  >  where  we  see  that  when 
7ve  divide  a  compound  quantity ^  we  divide  each  of  the  terms. 

§185.  We  must  also  recollect  that,  as  +  multiplied  by 
-f  J  makes  -f-  in  the  product,  so  +  in  the  product  divided 
by  +,  must  make  -f  in  the  quotient.  And  that  as  +  ^v^• 
tiplied  by  —  makes  — ,  so  —  in  the  product  divided  by  — 
will  bring  back  the  +  in  the  quotient.  So  that  when  the 
signs  are  alike  in  the  dividend  and  divisor,  the  sign  in  the 
quotient  is +.  Thus,  axb=ab;  both  of  which  are  +. 
Also — ax — b=ab  ;  and — ab-i 6=H-a. 

§186.  Again,  as  —  multiplied  by  +  makes  — ,  so  in 
the  product  —  divided  by  +,  brings  back  —  in  the  quotient. 
Also,  —  multiplied  by  —  makes  +  ;  and  of  course,  +  in 
the  product  divided  by  — ,  brings  —  in  the  quotient.  That 
is,  when  the  signs  in  the  divisor  and  dividend  are  unlike, 
the  sign  in  the  quotient  is  — . 

EXAMPLES. 

34.  Divide  2ad-\-8a^c  by  2a.  Ans.  d+Aac. 

35.  Divide  8d^7n^^l2d'm^  by  4(Zm».  Ans.  2d^—3d*m. 

36.  Divide  4.xy  +  6x^  by  2x.  Ans.  2y-^Sx. 

37.  Divide  abc — acd  by  ac  Ans.  b — d. 

38.  Divide  I2ax—8ab  by  —4a.  Ans.  — 3a;+26. 

39.  Divide  l0xz-{-15xy  by  5x. 

40.  Divide  15ax — 27iC  by  Sx, 

41.  Divide  18a^— 9a;  by  9a;. 


DIVISION  OF  ALGEBRAIC  QUANTITIES.  127 

42.  Divide  abc — bed — hex  by  — be. 

43.  Divide  3a?+6a^-|-3aa? — 15a;  by  3a:. 

44.  Divide  ^abe-\-V2abx—%a^b  by  3a6. 

45.  Divide  \Qa^b^-^e,Oa'b^—\lah  by  ab. 

46.  Divide  Iba^bc — \Octex^-\-^ad^c,  by — 5ac. 

47.  Divide  2(iax-\-lbax^-\-\Qax—ba,  by  5a. 

§187.  It  is  evident  that  we  may  divide  by  either  factor. 
Thus,  ax-\-bx  may  be  divided  by  a?,  and  the  quotient  will 
be  a+h ;  or  it  may  be  divided  by  a-\-b,  and  the  quotient 
will  be  x.  This  may  appear  singular  to  the  young  pupil ; 
but  he  is  to  recollect  that  division  is  merely  separating  the 
dividend  into  factors,  being  careful  to  rriake  one  of  them  of 
a  given  magnitude ;  that  is  to  make  it  the  same  as  the  given 
divisor. 

§188.  Now  we  know  that  x  times  a-\-b,=ax-{-bx  ;  and 
also  that  a-^b  times  x,=ax-\-bx.  Hence  we  know  that  the 
product  {ax-^bx)-^{a-\-h)=x.  Therefore  we  conclude  that 
if  the  divisor  contains  just  as  many  terms  as  the  dividend, 
with  corresponding  signs  ;  and  the  first  term  of  it  is  a  fac- 
tor in  the  first  term  of  the  dividend,  the  second  term  of  it  in 
the  second  of  the  dividend,  and  so  on  through  each  of  them 
respectively  ;  and  the  remaining  factor  in  each  term  of  the 
dividend  being  the  same  ;  that  remaining  factor  shall  be 
the  quotient. 

EXAMPLES. 

48.  Divide  ax+bx — ex,  by  a+b — c.  Ans.  x. 

49.  Divide  bac-{-bc^x — bx^,  by  ac-\-c^x — x^.  Ans.  b. 

50.  Divide  c'^ax — '2abx — 3xy  +  x,  by  ac^ — 2ab — 3?/+l. 

51.  Divide  cd^x—abd^'+d^x^—d'',  by  cx—ab+x^—1. 

52.  Divide  a^y — bcy-^xy,  by  a^ — bc  +  x. 

53.  Divide  6ahm — 14a&m — *icdm,  by  6ah — lAab — 3a/. 


I 


128  ALGEBRA. 

§189.  If  the  letters  of  the  divisor  are  nol  found  in  the 
dividend,  the  division  is  expressed,  as  we  have  before  shown, 
§177,  by  writing  the  divisor  underneath  the  dividend,  in  the 
form  of  a  vulgar  fraction. 

EXAMPLES. 

4i/"4~7.c 
64.  Divide  4y+7a;,  by  a — b.  Ans.— — —. 

55.  Divide  3a+262— ,c,  by  a+c, 

56.  Divide  a^— a^^ft  +  c^,  by  a^— fe^. 

57.  Divide  Sa''c+2b^  +  c,  by  2c, 

§190.  When  the  dividend  is  a  compound  quantity,  the 
divisor  may  be  placed  underneath  the  whole  dividend  if  we 
choose.  It  may  also  be  placed  under  each  term  of  the  di- 
vidend, which  is  the  same  as  dividing  each  term,  according 
to  §184.     By  this  method  the  answer  of  the  last  sum  would 

Sac^      2b^       c 
be 1- f-  — .     Answer  the  following  by  both  me- 

liC  'iC  <iC 

thods. 

58.  Divide  3a+6  +  2a6,  by  a.  Ans.  ~  +  5_+5f5. 

a       a        a 

59.  Divide  6a+a6— 3/^,  by  26. 

60.  Divide  2x-{-2y-\-^ax^2a^y,  by  Say. 

61.  Divide  fla;^— ftaj—a^^^+aft,  by  26. 

§191.  When  we  divide  each  term  separately,  we  may 
use  both  methods  of  division ;  that  is,  we  may  actually 
divide  such  terms  as  we  can  by  §180  ;  and  merely  express 
the  division  in  such  terms  as  cannot  be  divided. 

EXAMPLES. 

(IX 

62.  Divide  cd — ax-^ac+bc,  by  c.         Ans.  d h^+b. 

c 

63.  Divide ax+bx — 2ab-{-2x,  by  x.  Ans.  a+b \-2. 

X 


DIVISION  OF  ALGEBRAIC  QUANTITIES.  129 

64.  Divide  2am^Sa^b-\-b^m — Sa'^m,  by  — a, 

b^tn 

Ans.  — 2m  +  Sab \-Sam. 

a 

65.  Divide  2b*'^a''b-\-3b^c-^a^b^—ac,  by  6'. 

66.  Divide  ay^ — by^+4ta^b^ — 5a^by*-\-aby,  by  ay. 

67.  Divide  abx^+a^by — Sab^x-\-a3^ — 7a*i/,  by  ab. 

68.  Divide  2abm-irQa''b-^bb^m—4.a^m,  by  ab. 

69.  Divide  by — a^¥y-\-ay — aby-{-b^y,  by  —by. 

70.  Divide  ax — bx^-\-xy^-\-by — ay,  by  — y. 

In  all  of  our  divisions  so  far,  it  has  been  the  case  that 
either  the  divisor  or  the  quotient  is  a  simple  quantity.  For 
the  method  of  dividing  when  both  of  them  are  compound 
quantities^  see  §213,  &c. 

EXERCISES  IN  EQUATIONS. 

§192.  It  may  be  well  now  to  generalize  the  questions  in 

section  5  of  Equations.     And  with  this  section,   we  will 

make  our  calculations  more  purely  algebraical  than  in  the 

preceding  sections ;  as  we  shall  take  care  to  use  no  numeral 

quantities  at  all  in  stating  the  questions. 

1.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money 

in  trade.     A  gains  c  ($126,)  and  B  loses  d,{$&7  ;)  and  now 

A's  money  is  m  (two)  times  as  much  as  B's.     What  did 

each  lay  out?     See  page  45. 

Let     x=  what  each  lay  out. 

x-\-c=  A's  sum  now. 

X — c?=  B's  sum  now. 

mx — md=  m  times  B's. 

Forming  the  equation,  mu — md=^x-{-c 

Transposing,  mx — x~c+md 

^.  .,.      ,  c+md 

Dividmg  by  m — 1,  x= — —  . 

Substituting  numbers  ?  c+md       126+2x87  _ 


for  letters,  J  m — d  2 — 1 


130  ALGEBRA. 

In  sums  of  this  kind,  the  only  difficulty  is  to  determine 
what  quantity  to  divide  by  in  the  last  step,  to  leave  x  alone. 
But  this  difficulty  is  easily  overcome,  by  dividing  mentally 
the  left  hand  member  by  x,  and  observing  the  quotient. 
Of  course,  dividing  the  same  member  by  that  quotient  will 
produce  x ;  which  is  our  only  object. 

2.  The  2d  sum  on  page  45,  is  performed  as  follows : 
ip=z=  the  wife's  age. 
mx=  the  man's  age. 
ic+a=  wife's  after  a  years. 
mx-\-a=  man's  after  a  years. 
nx-}-na=  n  times  the  wife's  age. 
Forming  the  equation,        mx+a=^nx-]-na 
Transposing  terms,  nix — nx=na — a 


Dividing  by  m — n.  x= 


na- 


m — n 

Substituting  numbers,  aj=15,  the  age  of  his  wife. 

§193.  It  is  customary  to  represent  those  numbers  which 
stand  for  times,  by  the  letters  m,  n,  p,  q,  &c. 

All  the  other  questions  in  section  5th  are  to  be  performed 
in  the  foregoing  manner. 


DIVISION  OF  ALGEBRAIC  QUANTITIES.  131 


FRACTIONS. 
Reduction  of  Fractions  to  Lower  Terms. 

§194.  We  showed  in  §84,  that  a  fraction  may  be  re- 
duced to  lower  terms,  without  any  alteration  in  its  value, 
by  simply  dividing  both  terms  by  a  number  that  will  divide 
each  without  a  remainder.  Fractions  that  are  expressed 
by  literal  quantities  may  frequently  be  reduced  in  the  same 

GXV 

manner.     Thus,  in  the  fraction— r^,  both  the  terms  may  be 

divided  by  a  ;  and  the  fraction  will  then  become  7^. 

o 

EXAMPLES. 

A.(tbcx 
1.  Reduce  to  the  lowest  terms  the  fraction 


Qadcy 

,     ,  c  ^abcx      ^  2bx 

both  terms  }  - — ; i-2ac  =  -—7- ,     Ans. 

J  Gadcy  Sdy 

2.  Reduce    ,,    „    to  its  lowest  terms.  Ans.  -^  . 

a^om^x  abx 

^   ,       56ar^V®         .    ,  ,  8v^ 

3.  Reduce  — r-r— r  to  its  lowest  terms.     Ans. —, 

— Ix^y*  X* 

„    ,        — 4:X^yH      .     ,  ,  ^xH 

4.  Reduce  —-5 — ^—  to  its  lowest  terms.    Ans.  —  - — . 

ba^y*  52/2 

— 12x*i/z  3ic 

5.  Reduce  — t-t^  to  its  lowest  terms.         Ans.  —  . 

— ^x^yz*  z^ 

Oh^  These  examples  will  remind  the  pupil,  that,  (because 
fractions  are  merely  expressions  of  division,)  when  each 
term  has  its  sign,  then  the  whole  fraction  will  have  a  sign 
according  to  §185  and  186. 


132  ALGEBBA. 

D.  Keduce  to  its  lowest  terms. 

7.  Reduce  -^-^  to  its  lowest  terms. 

15aV5      .     , 

8.  Reduce  — to  its  lowest  terms. 

27amst 

§1 95.  When  we  divide  a  compound  quantity  by  a  simple 
quantity  we  divide  each  term,  §184.  Hence,  in  reducing 
fractions  to  lower  terms,  we  must  find  for  the  divisor,  a 
quantity  that  is  a  factor  in  every  term  both  of  the  numerator 
and  denominator. 

EXAMPLES. 

9.  Reduce  — — -=^  to  its  lowest  terms.     Ans.  — ~ . 

„    ,        a^oc^-{-ax^ — Sa*x       .     , 

10.  Reduce  — — .  ,  ,,  ^    to  its  lowest  terms. 

ax*  —  6ax^-}-9a^x 

aa^-\-ax — 3«3 

Ans   —————— ^^ 

x^—Qx-{-9a  ' 

,,    r.    1         4arv  —  8a?+12A       •     , 

11.  Reduce  — g — -— — -  to  its  lowest  terms. 

8ax^  —  4a^a? 

^„    ,        Sa^b^x  —  9lr^X'-6a3^      .       ' 

12.  Reduce  — —-r t^t—  to  its  lowest  terms. 

12aa;  +  6abx  —  I5bx 

„  _    ,        la^r^m  —  56am — 14am^       .     , 

1 3.  Reduce  --— — r^7i~j r^ ^o  its  lowest  terms. 

14.  Reduce  ^  ^  ^    .  o  t  «  ^ ;^  .  .      r.      to  its  lowest 

terms. 

^^  „   ,        112a6Ga:  —  48«C6Z+I00aca?      .    , 

15.  Reduce  — -—, — ; — ; — -— to  its  lowest  terms. 

^abcd  —  SacdJ^b2acx 

§196.  By  this  principle,  we  may  often  simplify  answers 
to  questions  in  division.     That  is,  we  may  put  the  divisor 


'  •'! 


FRACTIONS.  133 

underneath  the  dividend,  so  as  to  make  a  fraction,  and  then 
reduce  that  fraction  to  its  lowest  terms. 

EXAMPLES. 

16.  Divide  sP-  —  2xy-\-xif  by  4a-i/.  Ans.  ^~"  ^'^'^  , 

17.  Divide  10a?i/ — 20a? — 5?/ by — bx.  Ans. -, 

18.  Divide  7abx  —  bQa^xy+l^ax^  by  28a^bx^y. 

19.  Divide  Samy^-{-l6a^xy  —  24tabif  by  48ay  —  '72ay. 

20.  Divide  S5a^bc —  14aa?+42a^  by  21a'^  —  28a^x-^7a*x\ 

21.  Divide  S2x^i/+16x2f-^8xy^  by  24aa^+48a2a^. 

22.  Divide  5Aa^b^+^5a^b^  —  27a2i3  by  24a^6H30a365. 

23.  Divide  ^&x^y^  +  12axy  —  Wax  by  4a-y4-6aa!—  Sa^y. 

24.  Divide  28 (a  — 6+ a;)  by  4m(a— fe+a:.)    Ans.-. 

25.  Divide  I2cd{m  —  n)  by  14ac  (m  —  n), 

26.  Divide  6ah{r+p)  by  24«(r4-/)). 

27.  Divide  (tt-f  6)  (m+w)  by  (a  —  or)  (m-^n). 

§197.  It  must  be  remembered  that  when  all  the  factors 
in  the  numerator  are  contained  in  the  denominator,  the  an- 

Sa         1 

swer  will  contain  1  in  the  numerator.     Thus,  -- —  =  — . 

Sax      X 

MULTIPLICATION  WHERE  ONE  FACTOR  IS  A  FRACTION. 

§198.  This  is  done,  (as  shown  §69,  70,)  by  multiplying 
the  whole  number  and  the  numerator  of  the  fraction  toge- 

iher,  and  dividing  by  the  denominator.     Thus,  2a  x  — 

y 

dab  —  2ax 


12 


134  ALGEBRA. 


EXAMPLES. 


1.  Multiply  ^  by  3a.  Ans.  ?|^=  xy. 

2.  Multiply  - —  by  8a.  Ans.  — . 

^  ''  eax    ^  X 

3.  Multiply  3aa?  into  —  -^ — .  Ans.  —  — . 

4.  Multiply  2«6  -  3«,  by  |^.        Ans.  ""^'^  "  ^^. 

^    "^  30*  3aa; 

5.  Multiply  ^  by  20^  _  3^.       Ans.  !£%Z1^^^ 

CX  c 

6.  Multiply  Zam^  —  4a:  by  ?^!±£. 

7.  Multiply by  am  —  2a^. 

8.  Multiply  3a«y-f  6a:«  by  £?^JLf^. 

•^  3aa?  —  bd 

2ar* 
10.  Multiply  14a6c  —  3cdx  by       ^"^ 


bmx  —  2cm 


11.  Multiply ~ ^  by  .3mar  — 2am. 


ab 


12.  Multiply  ^  by  b,  Ans.-r  =  a. 

■      ,,  ,  .  ,    a6  ,  ^       aftc       a6 

13.  Mulfply^byc.  Ans.^  =  -. 

§^199.  In  the  last  example,  we  first  multiplied  the  nume- 
rator by  c  and  then  divided  both  the  numerator  and  the  de- 
nominator by  e.     Now,  multiplying  the  numerator  by  c 


amx 

Aus. 

8 

Ans. 

abc 
xy 

Ans. 

ax'y 

'    d   ' 

rRACTIONS.  185 

and  then  dividing  it  by  c,  is  altogether  useless ;  because 
the  numerator  is  left  as  it  was  at  first.  We  will  use  then 
only  one  part  of  the  operation ;  that  is,  dividing  the  deno- 
minator by  c.  And  in  general,  when  the  multiplier  is  a 
factor  in  the  denominator,  the  multiplication  is  performed 
by  canceling  that  factor* 

EXAMPLES. 

,  -    ,,  ,  .  ,    amx  . 

14.  Multiply by  r. 

15.  Multiply  ^  by  x'. 

16.  Multiply  1^  by  2cn. 

17.  Multiply  ^  by  2d. 

18.  Multiply  ^^  by  2x.         . 

19-  Multiply  ^g^"  by  36^. 
am^ 

»,    »r  ,  .  ,  6«m  —  4a;         ,     „ 

21.  Mu  tip  y —  by  3a. 

^  ^  3am+6aa?— 12a»    ^ 

^.    HT  1  •  ,  46c  —  bx+cx        .     „ 

oo    A*  1.-  1         am-^Aab  —  2m       ,      ^   , 
-'■  ''^"'''P'y  8abm+Wab-4abx  "^^  ""*• 

DIVISION  OF  FRACTIONS. 

§200.  We  have  shown  in  §95,  that  we  divide  fractions 
hy  dividing  the  numerator,  when  the  divisor  is  a  factor  in 


Ans» 

36 
x' 

Ans. 

Sax 

Ans. 

y^2 

26 

Ans.  ■ 

_xy  ^ 

3a  2 


136  ALGEBTRA. 

the  numerator  ;  hut  if  the  divisor  is  not  a  factor  in  the  nu- 
merator, we  multiply  the  denominator  by  the  divisor. 

EXAMPLES. 

I.Divide —  by  o. 

2.  Divide ^-r—  by  6. 
oc 

3.Divide5^^"  by  3x. 

26 

4.Divide^by  —  3a. 

z 

5.  Divide^  by  7aK 

6.  Divide  5^  by  7. 

^.  .,   3aa:  —  6ay  ,     ^ 

7.  Divide by  2a?y> 

4a  a?/ 

8.  Divide ; — -  by  3ay. 

a+2/ 

^.  .^   4a+27a6  .      . 

9.  Divide by  6aa,'. 

2x 

^.  .,   6a6ar  —  16axy  ^      ^ 
10. Divide-— — — - — -  by  2a, 
2am-\-Sxy 

^.  .,   146c^a?+21ac^^    ^  - 

11.  Divide — f byTc^ 

4a  6a; 

12.  Divide     ,  /i   „ 77; bySa^r*. 

a^4-5x^  —  lOaa; 

„  ^.  .,   12a6  —  I4ca:  ,     „  ., 

13. Divide—- —7—  by  2ac  —  Abx. 

10aarH-166c     '' 


FRACTIONS,  137 


1 4.  Divide ^: —  by  4ca  —  cax. 

x—Sy 

,_  ^.  .,   9am 4-1  Sax  ,     ,„  ^  „ 

15.  Divide -^ by  I2ax  —  9a'. 

6mx 

16.  Divide   ^^'"^^^^  by  a^  —  1. 

14a«— 16    ^ 


EXERCISES  IN  EQUATIONS. 

§201 .  Generalize  ihe  questions  in  Section  6th,  page  54. 
1.  In  an  orchard,  —  (i)  of  the  trees  bear  apples  i  —  (i) 

P 
of  them  bear  pears ;  —  (^-j-)  of  them  plums,  and  a  (81)  bear 

cherries.     How  many  trees  are  there  in  the  orchard? 

0:!?°  We  rarely  represent  unity  by  a  letter ;  but  general- 
ly use  its  own  character,  1. 

Let        X  =  number  of  trees. 

—  =  apple-trees. 

X 

— =  pear-trees. 
n      ^ 


J)X 

—   =  plum-trees. 


iheii,  X  —  — I [- —  +  a. 

771      71         r 

Multiplying  by  mnr,      innrx=^nrx-}-mrx  -\-mnpx+mnra 
Transposing,  mnrx  —  rirx  —  mrx  —  mnpx  =  mnra 

Dividing  by  mnr — nr — mr — mnp, 

mnra 

x= 

mn  r — n  r — mr — mnp 

Substitute  figures  for  letters,  and  find  the  answer. 
12* 


138  ALGEBRA. 

2d  sum,  page  55.     In  a  certain  school,  —  of  the  boys  learn 

mathematics  ;  —  of  them  study  Latin  and  Greek ;  and  a 
n  •' 

study  Grammar.     What  is  the  whole  number  of  scholars? 

The  equation  is,  ar  =  — [-  —  4-  a. 

Blultiplying  by  mn,  7nnx  =  na?  +  mpx  +  mna. 

Transposing,  mnx  —  nx  —  mpx  =  mna. 

r^.  .,.      ,  mna 

Dividmff  by  mn  —  n  —  m»,  x  = . 

°    "^  mn  —  n  —  mp 

3.  The  pupil  may  go  through  with  the  whole  section  in 
the  same  manner.  And  as  ih&xe  are  several  instances  in 
which  the  same  statement  and  the  same  answer  will  agree 
with  two  or  more  sums,  the  pupil  may  tell  which  they  are, 
and  why  it  so  happens. 

§202.  If  the  teacher  should  think  his  pupils  need  more 
practice,  he  may  exercise  them  in  the  7th  section  in  the 
same  manner. 

The  first  question  may  be  stated  thus, 

X  -^  a 

Multiplying  by  6,  a?  +  a  =  6c 

Transposing,  a?  =  6c  —  a. 

The  2d  will  have  the  following  equation. 

mx  4-  «_  , 

n     ~ 

Multiplying  by  n,  mx  -j- a  ^  nb 

Transposing  and  dividing,  _  nb  —  a 

m 

The  3d  and  5th  will  have  the  following  equation. 

nx  —  na 

m 

nx 
The  4th  will  be,  a  —  x=—. 

m 


FRACTIONS.  139 


FRACTIONS  OF  FRACTIONS. 

§203.  It  was  shown,  §96,  that  a  fraction  is  multiplied 
by  a  fraction^  by  multiplying  the  numerators  together  for 
a  new  numerator,  and  the  denominators  together  for  a  new 

denominator.     Thus,  -r- X -r=  r-f 
0       d      bd 

EXAMPLES. 

,,,,.,     3a  ^     4&  ,        I2ab 

1.  Multiply—  by—.  Ans.  — — . 

^  •'  5a;   "^    c  5cx 

^    ,,  ,  .  ,     9ax   ,     2bx  ,        leabx^      6x'' 

2.  Multiply  ^  by— .  Ans.j^^=-. 

3.  Find  the  product  of  ^i^  into  — ^.      Ans.-^^i^. 

^  Sa  X  —  y  ax — ay 

3x        4  a;  3a:^ 

4.  Multiply  —  and  —  together.  Ans.  -— . 

.    ,c  ,  •  ,      ^  .       6m  ,2 

.5.  Multiply—  into -^.  Ans.—. 

^  -^  Sy         7x        -  7 

^    ,,  ,  .  ,    3a  .  4a  ,  Sa^ 

6.  Multiply  -^  into  —  -  .  Ans.  —  -^ . 

7.  Multiply  linto-^.  Ans.  ^^±^  . 

^  -^  a  x+2y  ax-\-2ay 

a  .  y  .       ay 

5.  Multiply into —  — .  Ans.—. 

^    '  X  Z  XZ 

9.  Multiply  — ,  — ,  -^  together.        Ans.  — . 

4a  a?   3a?'M  2 

10.  What  is  the  product  of ,  -^,  and  — ?     Ans.  I2a?. 

, ,         ^     2a      .        2ac — be  ^ 

11.  What  IS  the  product  of     .         into 


36 +  c  Sab 


140  ALGEBRA. 

12.  Multiply  -j;^7+-2—  ^y  2S5;r3r5'c- 

13.  Multiply  ^Jg/  ty  g;^-^^^^- 

EXERCISES  IN  EQUATIONS. 

§204.  Generalize  the  questions  in  Equations,  Section  8. 

1.  The  1st  sum  on  page  6(5,  is  performed  as  follows. 
Stating  the  question,  x=  oats. 

mx      .     , 
— =  barley. 
n 

p       n       np 

mx  ,  mx 
Forming  the  equation,         *+~;^'^  "no  ~ 

Multiplying  by  np,         npx+mpx+mx  =  anp 

anp 
Dividing  by  np+mp-hm,       x  =  —j-^-j-^.       , 

X  X 

2.  The  equation  in  the  2d  sum  will  be  -—  — =  a 

Multiplying  by  mw,  mx  —  x^  amn 

__    amn 
Dividing  by  m  —  1,  ^  ~"  ^_i' 

3.  The  equation  in  the  3d  sum  will  be 

X         XX 

X =  «• 

m     n      mn 

4.  The  equation  in  the  4th  sum  will  be 

nx     nx     n^  __ 
m      m      m^ 
The  pupil  will  now  understand  how  to  perform  the  rest  of, 
the  section. 


FRACTIO>S.  ]  41 

UNITING  FRACTIONS  OF  DIFFERENT  DENOMINATIONS. 

§205.  Before  fractional  terms  can  be  united,  they  must 
be  brought  to  a  common  denominator,  according  to  the 
principle  explained  in  §81.  This  is  done  as  we  have  shown 
in  §102,  103,  by  multiplying  each  nvmerator  by  all  the 
denominators  except  its  own^for  new  numerators  ;  and  all 
the  denominators  together  for  a  new  denominator, 

EXAMPLES. 

,    TT  •.   .L    rn      •      .  2a6      ax  ,  am 

1 .  Unite  the  following  terms, 7/  "^  A~  * 

Sa¥x  —  abx^4-Aabmx      8ab'  —  ax^-\-Aa7n 

Ans. 77- = ~j . 

Ab^x^  Abx 

X  x  2x^ 

2.  Add  — -rrand together.      Ans.  -— ^ . 

o    C5  u.      *  3«  r        4a  .        32ac— 21a6 

6.  bubtract  -- from  — r.  Ans. — :-j . 

8c  lb  ^Qbc 

.    TT  .        3?  V  .       xz  —  xy  —  y^ 

4.  Unite  — •^.  '  Ans . . 

x-{-y       z  xz+yz 

_    ,^  .     X  —  y      x+y  .        ax^5ay 

5.  Unite      ^    -^ ~-^.  Ans. tt-o—^' 

2a  Sa  6a^ 

§206.  If,  in  such  cases,  there  is  a  quantity  that  is  not 
fractional,  we  multiply  it  by  all  the  denominators  ;  and  then 
putting  the  common  denominator  under  that  product,  unite 
it  with  the  other  fractions. 

^   TT  .        .  ^      A  ca         ,  ,       ea  ,  b      ac-\-b 

6.  Unite  a-\ — .  Ans.  a=—  ;  and  then,  — = . 

c  c  c      c  c 

-r   TT  •.        .  «+^  A        xy+a-^x 

7.  Unite  X-] —.  Ans.  -^ . 

y  y 

8.  Unite  a Ans. . 


142  ALGEBRA. 

ayz+8xz  —  '2x^y 


^   ,^  .      a   ,   4       ar     . 

0.  Unite  ^+2'+"-*. 

z            c 

-U'>"%-f-.+S-^ 

a                4 
2.  Unite xH-^ — a^/  — ;^- 

Ans. 

Ans. 


2xy2 


cx4-cy4-az —  bz 
Ans.  —  — 


Ans. 


cz 
2ay^+Sa^x+6ax 


2ay^+4y^ 
lbx-\-la  —  laby  —  46 


lb 

„    .,,2a;+l    4a?+2        ,  a?           ,  ,        169a:  +  77 

13.  Add  — - — ,  — - — ,  and  —  together.  Ans.  — -— — 

o           o                7  105 

,.    *.i          u      5a2+6        ,  4a2+26  ^        37a2+ll6 

14.  Add  together         ,    ,  and  — —r — .     Ans. —r — 

1^  cj  u.      .  ^^+1  r        21a;+3  ^       127a:+17 

15.  Subtract  — - —  from : .  Ans. 


28 

,«    CI,           3x4-1.        4a:  ^        /^x'^  —  UX'-b 

16.  Subtract ■—  from  — .  Ans 


x+l  5  5x4-5 

,-    Ajj  1.      2a:— 5       ^a:— 1     ^        4x3— .7a:  — 3 

1  / .  Add  together  — - — ,  and  —r — .  Ans. 


3     '  2a;  6a: 

18.  Add  together r,  and    — — .  Ans. 

°          X  —  3  a:+3  x^ — 9 

io  C!  K*      .2a:  — 3.       4a:42  ^       40:^^+3 

1 9.  Subtract  — from  — - — .  Ans. ^— 


3a:  3  3x 

20.  Add  together  — — j-,  and  — —7.  Ans. 


a  —  b  a  +  b  a^  —  6^ 

^^,    ^         3a426      ,  ^hd—^a—Zd 

^1.  rrom .subtract -, .  -,    , 

c  4crf  r  •<* 

12af/4-3iJ4-2a+3rf 

Ans. -—: ■ — . 

^cd 

22.  From  c42a6  —  3«c,  subtract  j- = — ■ —  . 

0^  —  be 

2ab^  —  6c^4  Sabc"  —  «' 


Ans. 


FRACTIONS.  143 


DIVISION  BY  FRACTIONS. 

§207.  Suppose  we  wish  to  know  how  many  times  -f-  is 
contained  in  ^.  We  would  divide  in  the  same  manner  that 
we  follow  in  dividing  6  pieces  by  3  pieces ;  and  say  ^  is 
contained  in  -f-,  two  times.  In  the  same  manner,  ^r  '^ 
contained  in  |^,  fae  times. 

The  principle  is  general  that  when  the  divisor  and  the 
dividend  have  a  common  denominator,  the  division  is  per • 
formed  by  dividing  the  numerator  of  the  dividend  by  the 
numerator  of  the  divisor, 

^,        a       c       a        a^      a^     a' 
Thus, -j-H-T-=  — ;      — ■7-—=— —a. 
b      b      c        X      X      a^ 

EXAMPLES. 

^    ^.  . ,    4a  ,     26  .       4a      2a 

1.  Divide  —  by  — .  Ans.--,  =  -r . 

^    ^.  .,     Sab  ,     Abx  ,       Sab      Sa 

2.  Divide    —7-  by  — r-  Ans.— r--=  -— . 

cd     •'    cd  Abx     Ax 

3.  Divide  ~r  by  —r*  Ans.  •—  . 

ab     ^  ab  Sy 

4.  Divide T—    by  -t~.  Ans.     ^  -    . 

ab        ^  ab  2a^ 


I 


^    ^.  .J    7raT  +  a«  ^    Ab+ax  lrx-{-a'' 

5.  Divide  — - — -—  by  —= — -—       Ans.-y- 

bast  5u8t  46+oa* 

^    T..  ..    ^    .     2a6 

6.  Divide  8a  by . 

Explanation,     By  §206,  6a  =  -^.      Then  — ^ 

8a6  _  8axy  __  Aocy 
xy         2ab         b 


144  ALGEBRA. 

7.  Divide  Sa  by  —.  Ans.  — -. 

U  4a; 

^    ,^.  . ,      ,  ,     c  ^       abd 

8.  Divide  ao  by  —.  Ans.  — . 

a  c 

9.  Divide  by .  Ans.  — -  . 

y          y  x^ 

10.  Divide  by  .  Ans. . 

oa?       ''ax  Sam 

^.  .,     4ay^x  ,     lOaa;'^  .       2i/'' 

11.  Divide  —f- —   by — r — .  Ans.--^. 

oc              be  5x 

12.  Divide by .  Ans 


ax        -^       ax  4a*  —  ab' 

13.  Diviae  -; by-; .  Ans. — . 

ab  —  c     ''  ab  —  c  x 

,  ^    T^.  . ,     ax  —  ly        xy 

14.  Divide  j-^  by  -^. 

ab         ''    b 

Explanation,  In  this  example,  the  dividend  and  the  di- 
visor has  not  a  common  denominator.  Our  first  object 
then,  is  to  bring  them  to  a  common  denominator.     This  is 

,        ,      {.•r^^   ^^  —  V       xy       ^ax  —  by       abxy 

done  by  6^5, —^  -^  •#  = rr-^  -^  — i/. 

^  ^  ab  b  ab'^  ab^ 

abx  —  by       ax  —  ?/ 
Ans.  ^  —  '' 


Id.  Divide  by -^. 

c       ^  a 


abxy  axy 

abx       (ly       a"bx       cily       a^bx 


Operation.  .  .  —      ,    . 

c  a  ac         ae         edy 

§208.  It  will  be  seen  that  in  sums  of  this  kind,  after  we 
have  brought  the  terms  to  a  common  denominator,  the  divi- 
sion is  performed  by  putting  the  numerator  of  the  dividend 
for  the  numerator  of  the  answer,  and  the  numerator  of  the 


FBACTIONS.  145 

divisor  for  the  denominator  of  the  answer,  and  make  no  use 
at  all  of  the  denominators.  Let  us  see  then  how  we  obtain 
these  two  terms.  We  multiply  the  numerator  of  the  divi- 
dend by  the  denominator  of  the  divisor ;  and  this  becomes 
the  numerator  of  the  answer.  And  we  miultiply  the  nume- 
rator of  the  divisor  by  the  denominator  of  the  dividend ; 
and  this  becomes  the  denominator  of  the  answer.  By  look- 
ing at  the  last  two  sums,  it  will  be  seen  that  this  is  the  true 
operation. 

§209.  Hence  we  obtain  the  general  rule  for  dividing  by 
a  fraction.  Multiply  the  numerator  of  the  dividend  by  the 
denominator  of  the  divisor,  for  a  new  numerator ;  and 
multiply  the  denominator  of  the  dividend  by  the  numerator 
of  the  divisor  for  a  new  denominator. 

§210.  When  the  dividend  is  a  whole  number,  it  is  chang- 
ed into  a  fraction,  by  putting  1  under  it  for  a  denominator. 


Thus,^  =  ^;         2a  =   5^. 

EXAMPLES. 

16. 

Divide-  by  ^^.             Ans.  ^^^=2^. 

17. 

^.  .,    a+x  ,      o+V       A       a^+'6ax-\-2x^ 
Divide   -!—  by  —ff--     Ans.  —4 — ^ — . 
a— y    ^  a-\-2x                   a'^  —  y^ 

18. 

T^i'virlr    ^^        "/yy-t-iT^   hv 

jjiviae  X  ■— "CuXT^tt     uy              • 
X  —  a 

Ans.  a?3  —  dx^a+Sa'^x  —  a 

19. 

Divide  -by-.              Ans.-. 

20. 

^.  .,     x  —  l  ,    x+l     ,       4a;  — 4 
Divide       3       by     ^    .   Ans.  ^^  _^  3. 

13 

146^  ALGEBRA. 


01     r^-  -A    "+^  u    ^"  A       4a6+4&r 

21.  Divide  by  T/-  Ans.  — . 

y  4o  day 

22.  Divide  — r-   by Ans.  —  --#. 

4       -^         y  16 

23.  Divide  4-  by  ??.  Ans. -i^.  . 

a+1     "^  41/  3tta;+3a? 

24.  Divide  a;+aa;  by .  Ans.-—!- ^ i. 

•'a?  —  y  Sa 

25.  Divide  -^—  by  |-.         Ans. ^. 

26.  Divide  4a  —  av  by -^^ -.     Ans.  -• 

^    ''      A^a  4y  —  ay 

27.  Divide  by  — „.         Ans.  x-{-y. 

X  —  y     ^  x^ — y^  ^ 

REDUCTION  OF  COMPLEX  FRACTIONS  TO  SIMPLE  ONES. 

§211.  We  have  shown,  that  when  we  multiply  a 
fraction  by  its  denominator,  we  obtain  for  the  answer,  the 
same  quantity  as  the  numerator.  We  have  also  shown  that 
where  both  terms  of  a  fraction  are  muhiplied  by  the  same 
quantity,  the  value  of  the  fraction  is  not  altered.  By  these 
two  principles,  we  obtain  the  following  rule  for  reducing  a 
complex  fraction  to  a  simple  one. 

§212.  Multiply  both  terms  of  the  fraction  by  the  deno- 
minator that  is  found  either  in  the  entire  numerator  or  de- 
nominator. If  the  fraction  is  still  complex^  multiply  the 
result  in  both  terms  by  the  remaining  denominator  that  is 
found  in  the  entire  term.     Thus, 

a+A      ca+b^          a  — f      ca  ^  b       4ac— 46 


cd 


3      ,    ^   —    3c 


4+a       ^  +  ca       3c+4ac 
4  4    •  • 


FRACTIONS.  147 

EXAMPLES. 

1.  Reduce r-  to  a  simple  fraction. 

c 

ac 


Ans.  Multiplying  both  terms  by  c, 


ac  —  b' 


2.  Reduce  a  • —  iTto  a  simple  fraction.  Ans.  — . 

ax  ^ 

X 

3.  Reduce to  a  simple  fraction. 

ay  +  ^  ^ 

5x 
Ans 


^^y+y  —  z 

4:X — X         3 


4.  Reduce  ^        4  to  a  simple  fraction.  Ans     , 

5.  Reduce  - — -  ,,_i  to  a  simple  fraction.  Ans. ^— -. . 

3a  — =^  ^  3ay— 1/+1 

X 

6.  Reduce  ^  +  y  to  a  simple  fraction. 

z  —  ± 

e 

Ans. y  _  xy  -\-  x       cxy  -f-  ex 

z — JL      yz — y^    ~'cyz  —  y^ 
c 

£.  24z  +  6x 

7.  Reduce  4  +  z   to  a  simple  fraction.      Ans. — . 

I  VXZ  —  OZ 

X — ± 
6 

X  cxv^  "4"  ex 

8.  Reduce  ^y  +  ^  to  a  simple  fraction.     Ans.  — ^  .  ,   .,. 

^TfTT  '  acy-\-bxy 


148  AL6EBRA. 


9.  Reduce  3  4-^    to  a  simple  fraction.     Ans.  — ^^- — - 


4  — ± 


Adcy — bx 


oy  hcx~~~'CiC'U 

JO.  Reduce  x —  6  to  a  simple  fraction.       Ans.  -, r^* 

^Zf^  bcy+abx 

c 

§213.  It  sometimes  happens  that  we  wish  to  transfer  a  frac- 
tion from  a  numerator  to  the  denominator,  or  from  a  denomi- 
nator to  the  numerator.    This  may  be  done  by  the  foregoing 

principles.     For,  supposing  we  have  ~ ;  multiplying  by  the 

denominator  of  f ,  we  have  — .      Now,  if  we   divide  this 
/  X 

CL 

fraction  by  2,  we  have  —  .     Thus  we  see   the  fraction  is 

transferred,  without  altering  the  value  of  the  whole  quantity, 
if  we  take  care  to  invert  it  when  we  transfer  it. 


THE  END. 


